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Mental calculation
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This article contains instructions, advice, or how-to content. The purpose of Wikipedia is to present facts, not to train. Please help improve this article either by rewriting the how-to content or by moving it to Wikiversity, Wikibooks or Wikivoyage. (February 2017)
Mental calculation comprises arithmetical calculations using only the human brain, with no help from any supplies (such as pencil and paper) or devices such as a calculator. People use mental calculation when computing tools are not available, when it is faster than other means of calculation (such as conventional educational institution methods), or even in a competitive context. Mental calculation often involves the use of specific techniques devised for specific types of problems.
Many of these techniques take advantage of or rely on the decimal numeral system. Usually, the choice of radix is what determines which method or methods to use.
Contents
[hide]
1 Methods and techniques
1.1 Casting out nines
1.2 Estimation
1.3 Factors
1.4 Calculating differences: a − b
1.4.1 Direct calculation
1.4.2 Indirect calculation
1.4.3 Look-ahead borrow method
1.5 Calculating products: a × b
1.5.1 Multiplying any two numbers by attaching, subtracting, and routing
1.5.1.1 The "Ends of Five" Formula
1.5.1.2 The "Borrower's" Formula
1.5.2 Multiplying any 2-digit numbers
1.5.3 Multiplying by 2 or other small numbers
1.5.4 Multiplying by 5
1.5.5 Multiplying by 9
1.5.5.1 Using hands: 1–10 multiplied by 9
1.5.6 Multiplying by 10 (and powers of ten)
1.5.7 Multiplying by 11
1.5.8 Multiplying two 2 digit numbers between 11 and 19
1.5.9 Using hands: 6–10 multiplied by another number 6–10
1.5.10 Multiplying two numbers close and below 100
1.6 Using square numbers
1.7 Squaring numbers
1.7.1 Squaring any number
1.7.2 Squaring any 2-digit integers
1.7.3 Squaring a number ending in 5
1.7.4 Squaring numbers near 50
1.7.5 Squaring an integer from 26 to 75
1.7.6 Squaring an integer from 76 to 125
1.8 Finding roots
1.8.1 Approximating square roots
1.8.2 Extracting roots of perfect powers
1.8.2.1 Extracting cube roots
1.9 Approximating common logs (log base 10)
1.10 Approximating natural exponents (log base e)
1.10.1 Table of approximations
1.10.2 Optimized Relations
1.10.3 Physical Science and Communication Approximations
1.10.4 Other examples
2 Mental arithmetic as a psychological skill
3 Mental Calculations World Championship
4 Mental Calculation World Cup
5 Memoriad – World Memory, Mental Calculation & Speed Reading Olympics
6 See also
7 References
8 External links
Methods and techniques[edit]
Casting out nines[edit]
Main article: Casting out nines
After applying an arithmetic operation to two operands and getting a result, you can use this procedure to improve your confidence that the result is correct.
Sum the digits of the first operand; any 9s (or sets of digits that add to 9) can be counted as 0.
If the resulting sum has two or more digits, sum those digits as in step one; repeat this step until the resulting sum has only one digit.
Repeat steps one and two with the second operand. You now have two one-digit numbers, one condensed from the first operand and the other condensed from the second operand. (These one-digit numbers are also the remainders you would end up with if you divided the original operands by 9; mathematically speaking, they're the original operands modulo 9.)
Apply the originally specified operation to the two condensed operands, and then apply the summing-of-digits procedure to the result of the operation.
Sum the digits of the result you originally obtained for the original calculation..
If the result of step 4 does not equal the result of step 5, then the original answer is wrong. If the two results match, then the original answer may be right, though it isn't guaranteed to be.
Example
Say we've calculated that 6338 × 79 equals 500702
Sum the digits of 6338: (6 + 3 = 9, so count that as 0) + 3 + 8 = 11
Iterate as needed: 1 + 1 = 2
Sum the digits of 79: 7 + (9 counted as 0) = 7
Perform the original operation on the condensed operands, and sum digits: 2 × 7 = 14; 1 + 4 = 5
Sum the digits of 500702: 5 + 0 + 0 + (7 + 0 + 2 = 9, which counts as 0) = 5
5 = 5, so there's a good chance that we were right that 6338 × 79 equals 500702.
You can use the same procedure with multiple operations just repeat steps 1 and 2 for each operation.
Estimation[edit]
While checking the mental calculation, it is useful to think of it in terms of scaling. For example, when dealing with large numbers, say 1531 × 19625, estimation instructs you to be aware of the number of digits expected for the final value. A useful way of checking is to estimate. 1531 is around 1500, and 19625 is around 20000, so a result of around 20000 × 1500 (30000000) would be a good estimate for the actual answer (30045875). So if the answer has too many digits, you know you've made a mistake.
Factors[edit]
When multiplying, a useful thing to remember is that the factors of the operands still remain. For example, to say that 14 × 15 was 211 would be unreasonable. Since 15 is a multiple of 5, the product should be as well. Likewise, 14 is a multiple of 2, so the product should be even. Furthermore, any number which is a multiple of both 5 and 2 is necessarily a multiple of 10, and in the decimal system would end with a 0. The correct answer is 210. It is a multiple of 10, 7 (the other prime factor of 14) and 3 (the other prime factor of 15).
Calculating differences: a − b[edit]
Direct calculation[edit]
When the digits of b are all smaller than the corresponding digits of a, the calculation can be done digit by digit. For example, evaluate 872 − 41 simply by subtracting 1 from 2 in the units place, and 4 from 7 in the tens place: 831.
Indirect calculation[edit]
When the above situation does not apply, the problem can sometimes be modified:
If only one digit in b is larger than its corresponding digit in a, diminish the offending digit in b until it is equal to its corresponding digit in a. Then subtract further the amount b was diminished by from a. For example, to calculate 872 − 92, turn the problem into 872 − 72 = 800. Then subtract 20 from 800: 780.
If more than one digit in b is larger than its corresponding digit in a, it may be easier to find how much must be added to b to get a. For example, to calculate 8192 − 732, we can add 8 to 732 (resulting in 740), then add 60 (to get 800), then 200 (for 1000). Next, add 192 to arrive at 1192, and, finally, add 7000 to get 8192. Our final answer is 7460.
Another technique we can use is rounding off one of the digits (Either the larger digit or the smaller digit to the nearest number preferably containing a single non-zero digit). For example, to calculate 8192 - 732 we can round off 732 to 1000 by adding 268 (We can find the value 268 by quickly subtracting 732 from 1000 mentally. Our brain finds it easier in dealing with rounded figures.) to it. Then we can subtract 1000 from 8192 and obtain 7192 as the answer. Now we can add 268 to 7192 to obtain 7460 as the answer.
Alternatively we can exchange numbers to round off figures as in the given problem. For example, For calculating 8192 - 732 we can simply add 268 to both sides resulting in 8460 - 1000, which is easier to calculate, resulting in 7460.
One needs to be prudent in selecting which number to be rounded off. For calculating 8192 - 732 we can round off 8192 to 9000 by adding 808. Then we calculate 9000-732 resulting in 8268. Then we can subtract 808 from 8268 to get 7460 as our answer. But as we can observe, we made the calculations difficult and lengthy.
We may also perform the calculation in the traditional way but in a smart manner. For calculating 8192 - 732 we can eliminate the 2 in units place i.e. replace them with 0. Then we can subtract 3 from 9 yielding in 6. Finally we can subtract 7 from 81 yielding in 74. Then we just need to rearrange the pieces to get 7460 as the answer
It might be easier to start from the left (the big numbers) first.
You may guess what is needed, and accumulate your guesses. Your guess is good as long as you haven't gone beyond the "target" number. 8192 − 732, mentally, you want to add 8000 but that would be too much, so we add 7000, then 700 to 1100, is 400 (so far we have 7400), and 32 to 92 can easily be recognized as 60. The result is 7460.
Look-ahead borrow method[edit]
This method can be used to subtract numbers left to right, and if all that is required is to read the result aloud, it requires little of the user's memory even to subtract numbers of arbitrary size.
One place at a time is handled, left to right.
Example:

          4075
        − 1844
        ------

Thousands: 4 − 1 = 3, look to right, 075 < 844, need to borrow.
           3 − 1 = 2, say "Two thousand".
           We are performing 3 - 1 rather than 4 - 1 because the column to the right is
           going to borrow from the thousands place.

Hundreds: 0 − 8 = negative numbers not allowed here.
          We are going to increase this place by using the number we borrowed from the
          column to the left. Therefore:
          10 − 8 = 2. It's 10 rather than 0, because we borrowed from the Thousands
          place. 75 > 44 so no need to borrow,
          say "two hundred"

Tens: 7 − 4 = 3, 5 > 4, so 5-4=1

Hence the result is 2231
Calculating products: a × b[edit]
Many of these methods work because of the distributive property.
Multiplying any two numbers by attaching, subtracting, and routing[edit]
Discovered by Artem Cheprasov, there is a method of multiplication that allows the user to utilize 3 steps to quickly multiply numbers of any size to one another via three unique ways.[1][2]
First, the method allows the user to attach numbers to one another, as opposed to adding or subtracting them, during intermediate steps in order to quicken the rate of multiplication. For instance, instead of adding or subtracting intermediary results such as 357 and 84, the user could simply attach the numbers together (35784) in order to simplify and expedite the multiplication problem. Attaching numbers to one another helps to bypass unnecessary steps found in traditional multiplication techniques.
Secondly, this method uses negative numbers as necessary, even when multiplying two positive integers, in order to quicken the rate of multiplication via subtraction. This means two positive integers can be multiplied together to get negative intermediate steps, yet still the correct positive answer in the end. These negative numbers are actually automatically derived from the multiplication steps themselves and are thus unique to a particular problem. Again, such negative intermediate steps are designed to help hasten the mental math.
Finally, another unique aspect of using this method is that the user is able to choose one of several different “routes of multiplication” to the specific multiplication problem at hand based on their subjective preferences or strengths and weaknesses with particular integers.
Despite the same starting integers, the different multiplication routes give off different intermediate numbers that are automatically derived for the user as they multiply. Some of these intermediaries may be easier than others (e.g. some users may find a route that uses a negative 7, while another route uses a 5 or a 0, which are typically easier to work with mentally for most people, but not in all instances).
If one “route” seems to be harder for one student vs. another route and its intermediate numbers, that student can simply choose another simpler route of multiplication for themselves even though it’s the same original problem.
The "Ends of Five" Formula[edit]
For any 2 digit by 2 digit multiplication problem, if both numbers end in five, the following algorithm can be used to quickly multiply them together:[1]
E x : 35 × 75 {\displaystyle Ex:35\times 75}
As a preliminary step simply round the smaller number down and the larger up to the nearest multiple of ten. In this case:
35 − 5 = 30 = X {\textstyle 35-5=30=X}
75 + 5 = 80 = Y {\displaystyle 75+5=80=Y}
The algorithm reads as follows:
( X × Y ) + 50 ( t 1 − t 2 ) + 25 {\displaystyle (X\times Y)+50(t_{1}-t_{2})+25}
Where t1 is the tens unit of the original larger number (75) and t2 is the tens unit of the original smaller number (35).
= 30 × 80 + 50 ( 7 − 3 ) + 25 = 2625 {\displaystyle =30\times 80+50(7-3)+25=2625}
The author also outlines another similar algorithm if you want to round the original larger number down and the original smaller number up instead.
The "Borrower's" Formula[edit]
If two numbers are equidistant from the nearest multiple of 100, then a simple algorithm can be used to find the product.[1]
As a simple example:
33 × 67 {\displaystyle 33\times 67}
Both numbers are equidistant (33 away) from their nearest multiple of 100 (0 and 100, respectively).
As a preliminary step simply round the smaller number down and the larger up to the nearest multiple of ten. In this case:
33 − 3 = 30 = X {\textstyle 33-3=30=X}
67 + 3 = 70 = Y {\displaystyle 67+3=70=Y}
The algorithm reads as follows:
( X × Y ) + u 1 × u 2 + u 2 ( T 1 − T 2 ) {\displaystyle (X\times Y)+u_{1}\times u_{2}+u_{2}(T_{1}-T_{2})}
Where u1 is the original larger number's (67) units digit and u2 is the original smaller number's (33) units digit. T1 is the original larger number's tens digit and T2 is the original larger number's tens digit multiplied by their respective power (in this case by 10, for a tens digit).
And so:
( 30 × 70 ) + 7 × 3 + 3 ( 60 − 30 ) = 2100 + 21 + 90 = 2211 {\displaystyle (30\times 70)+7\times 3+3(60-30)=2100+21+90=2211}
Multiplying any 2-digit numbers[edit]
To easily multiply any 2-digit numbers together a simple algorithm is as follows (where a is the tens digit of the first number, b is the ones digit of the first number, c is the tens digit of the second number and d is the ones digit of the second number):
( 10 a + b ) ⋅ ( 10 c + d ) {\displaystyle (10a+b)\cdot (10c+d)}
= 100 ( a ⋅ c ) + 10 ( b ⋅ c ) + 10 ( a ⋅ d ) + b ⋅ d {\displaystyle =100(a\cdot c)+10(b\cdot c)+10(a\cdot d)+b\cdot d}
For example,
23 ⋅ 47 = 100 ( 2 ⋅ 4 ) + 10 ( 3 ⋅ 4 ) + 10 ( 2 ⋅ 7 ) + 3 ⋅ 7 {\displaystyle 23\cdot 47=100(2\cdot 4)+10(3\cdot 4)+10(2\cdot 7)+3\cdot 7}
  800
 +120
 +140
 + 21
-----
 1081

Note that this is the same thing as the conventional sum of partial products, just restated with brevity. To minimize the number of elements being retained in one's memory, it may be convenient to perform the sum of the "cross" multiplication product first, and then add the other two elements:
( a ⋅ d + b ⋅ c ) ⋅ 10 {\displaystyle (a\cdot d+b\cdot c)\cdot 10}
+ b ⋅ d {\displaystyle {}+b\cdot d} [of which only the tens digit will interfere with the first term]
+ a ⋅ c ⋅ 100 {\displaystyle {}+a\cdot c\cdot 100}
i.e., in this example
(12 + 14) = 26, 26 × 10 = 260,
to which is it is easy to add 21: 281 and then 800: 1081
An easy mnemonic to remember for this would be FOIL. F meaning first, O meaning outer, I meaning inner and L meaning last. For example:
75 ⋅ 23 {\displaystyle 75\cdot 23}
and
a b ⋅ c d {\displaystyle ab\cdot cd}
where 7 is a, 5 is b, 2 is c and 3 is d.
Consider
a ⋅ c ⋅ 100 + ( a ⋅ d + b ⋅ c ) ⋅ 10 + b ⋅ d {\displaystyle a\cdot c\cdot 100+(a\cdot d+b\cdot c)\cdot 10+b\cdot d}
this expression is analogous to any number in base 10 with a hundreds, tens and ones place. FOIL can also be looked at as a number with F being the hundreds, OI being the tens and L being the ones.
a ⋅ c {\displaystyle a\cdot c} is the product of the first digit of each of the two numbers; F.
( a ⋅ d + b ⋅ c ) {\displaystyle (a\cdot d+b\cdot c)} is the addition of the product of the outer digits and the inner digits; OI.
b ⋅ d {\displaystyle b\cdot d} is the product of the last digit of each of the two numbers; L.
Multiplying by 2 or other small numbers[edit]
Where one number being multiplied is sufficiently small to be multiplied with ease by any single digit, the product can be calculated easily digit by digit from right to left. This is particularly easy for multiplication by 2 since the carry digit cannot be more than 1.
For example, to calculate 2 × 167: 2×7=14, so the final digit is 4, with a 1 carried and added to the 2×6 = 12 to give 13, so the next digit is 3 with a 1 carried and added to the 2×1=2 to give 3. Thus, the product is 334.
Multiplying by 5[edit]
To multiply a number by 5,
1. First multiply that number by 10, then divide it by 2. The two steps are interchangeable i.e. you can halve the number and then multiply it.
The following algorithm is a quick way to produce this result:
2. Add a zero to right side of the desired number. (A.) 3. Next, starting from the leftmost numeral, divide by 2 (B.) and append each result in the respective order to form a new number;(fraction answers should be rounded down to the nearest whole number).
EXAMPLE: Multiply 176 by 5.
     A. Add a zero to 176 to make 1760.
     B. Divide by 2 starting at the left.
           1. Divide 1 by 2 to get .5, rounded down to zero.
           2. Divide 7 by 2 to get 3.5, rounded down to 3.
           3. Divide 6 by 2 to get 3. Zero divided by two is simply zero.

The resulting number is 0330. (This is not the final answer, but a first approximation which will be adjusted in the following step:)
     C. Add 5 to the number that follows any single numeral
        in this new number that was odd before dividing by two;

EXAMPLE: 176 (IN FIRST, SECOND THIRD PLACES):
           1.The FIRST place is 1, which is odd. ADD 5 to the numeral after
             the first place in our new number (0330) which is 3; 3+5=8.

           2.The number in the second place of 176, 7, is also odd. The
             corresponding number (0 8 3 0) is increased by 5 as well;
             3+5=8.

           3.The numeral in the third place of 176, 6, is even, therefore
             the final number, zero, in our answer is not changed. That
             final answer is 0880.
             The leftmost zero can be omitted, leaving 880.
             So 176 times 5 equals 880.

EXAMPLE: Multiply 288 by 5.
A. Divide 288 by 2. We can divide each digit individually to get 144. (Dividing smaller number is easier.)
B. Multiply by 10. Add a zero to yield the result 1440.
Multiplying by 9[edit]
Since 9 = 10 − 1, to multiply a number by nine, multiply it by 10 and then subtract the original number from the result. For example, 9 × 27 = 270 − 27 = 243.
This method can be adjusted to multiply by eight instead of nine, by doubling the number being subtracted; 8 × 27 = 270 − (2×27) = 270 − 54 = 216.
Similarly, by adding instead of subtracting, the same methods can be used to multiply by 11 and 12, respectively (although simpler methods to multiply by 11 exist).
Using hands: 1–10 multiplied by 9[edit]
Hold hands in front of you, palms facing you. Assign the left thumb to be 1, the left index to be 2, and so on all the way to right thumb is ten. Each "|" symbolizes a raised finger and a "−" represents a bent finger.
1 2 3 4 5  6 7 8 9 10
| | | | |  | | | | |
left hand  right hand

Bend the finger which represents the number to be multiplied by nine down.
Ex: 6 × 9 would be
| | | | |  − | | | |

The right little finger is down. Take the number of fingers still raised to the left of the bent finger and prepend it to the number of fingers to the right.
Ex: There are five fingers left of the right little finger and four to the right of the right little finger. So 6 × 9 = 54.
    5           4
| | | | |  − | | | |

Multiplying by 10 (and powers of ten)[edit]
To multiply an integer by 10, simply add an extra 0 to the end of the number. To multiply a non-integer by 10, move the decimal point to the right one digit.
In general for base ten, to multiply by 10n (where n is an integer), move the decimal point n digits to the right. If n is negative, move the decimal |n| digits to the left.
Multiplying by 11[edit]
For single digit numbers simply duplicate the number into the tens digit, for example: 1 × 11 = 11, 2 × 11 = 22, up to 9 × 11 = 99.
The product for any larger non-zero integer can be found by a series of additions to each of its digits from right to left, two at a time.
First take the ones digit and copy that to the temporary result. Next, starting with the ones digit of the multiplier, add each digit to the digit to its left. Each sum is then added to the left of the result, in front of all others. If a number sums to 10 or higher take the tens digit, which will always be 1, and carry it over to the next addition. Finally copy the multipliers left-most (highest valued) digit to the front of the result, adding in the carried 1 if necessary, to get the final product.
In the case of a negative 11, multiplier, or both apply the sign to the final product as per normal multiplication of the two numbers.
A step-by-step example of 759 × 11:
The ones digit of the multiplier, 9, is copied to the temporary result.
result: 9
Add 5 + 9 = 14 so 4 is placed on the left side of the result and carry the 1.
result: 49
Similarly add 7 + 5 = 12, then add the carried 1 to get 13. Place 3 to the result and carry the 1.
result: 349
Add the carried 1 to the highest valued digit in the multiplier, 7 + 1 = 8, and copy to the result to finish.
Final product of 759 × 11: 8349
Further examples:
−54 × −11 = 5 5+4(9) 4 = 594
999 × 11 = 9+1(10) 9+9+1(9) 9+9(8) 9 = 10989
Note the handling of 9+1 as the highest valued digit.
−3478 × 11 = 3 3+4+1(8) 4+7+1(2) 7+8(5) 8 = −38258
62473 × 11 = 6 6+2(8) 2+4+1(7) 4+7+1(2) 7+3(0) 3 = 687203
Another method is to simply multiply the number by 10, and add the original number to the result.
For example:
17 × 11
     17 × 10 = 170

170 + 17 = 187
17 × 11 = 187
One last easy way:
If you have a two-digit number, take it and add the two numbers together and put that sum in the middle, and you can get the answer.
For example: 24 x 11 = 264 because 2 + 4 = 6 and the 6 is placed in between the 2 and the 4.
Second example: 87 x 11 = 957 because 8 + 7 = 15 so the 5 goes in between the 8 and the 7 and the 1 is carried to the 8. So it is basically 857 + 100 = 957.
Or if 43 x 11 is equal to first 4+3=7 (For the tens digit) Then 4 is for the hundreds and 3 is for the tens. And the answer is 473
Multiplying two 2 digit numbers between 11 and 19[edit]
To easily multiply 2 digit numbers together between 11 and 19 a simple algorithm is as follows (where a is the ones digit of the first number and b is the ones digit of the second number):
(10+a)×(10+b)
100 + 10×(a+b) + a×b

which can be visualized as three parts to be added:

1
xx
 yy

for example:

17×16

1               = 100
13    (7+6)     = 10×(a+b)
 42   (7×6)     = a×b
272   (total)

Using hands: 6–10 multiplied by another number 6–10[edit]
This technique allows a number from 6 to 10 to be multiplied by another number from 6 to 10.
Assign 6 to the little finger, 7 to the ring finger, 8 to the middle finger, 9 to the index finger, and 10 to the thumb. Touch the two desired numbers together. The point of contact and below is considered the "bottom" section and everything above the two fingers that are touching are part of the "top" section. The answer is formed by adding ten times the total number of "bottom" fingers to the product of the number of left- and right-hand "top" fingers.
For example, 9 × 6 would look like this, with the left index finger touching the right little finger:
                               =10==   :right thumb              (top)
                               ==9==   :right index finger       (top)
                               ==8==   :right middle finger      (top)
        left thumb:   =10==    ==7==   :right ring finger        (top)
 left index finger:   --9---><---6--   :right little finger      (BOTTOM)
left middle finger:   --8--                                      (BOTTOM)
  left ring finger:   --7--                                      (BOTTOM)
left little finger:   --6--                                      (BOTTOM)

In this example, there are 5 "bottom" fingers (the left index, middle, ring, and little fingers, plus the right little finger), 1 left "top" finger (the left thumb), and 4 right "top" fingers (the right thumb, index finger, middle finger, and ring finger). So the computation goes as follows: 9 × 6 = (10 × 5) + (1 × 4) = 54.
Consider another example, 8 × 7:
                               =10==   :right thumb              (top)
        left thumb:   =10==    ==9==   :right index finger       (top)
 left index finger:   ==9==    ==8==   :right middle finger      (top)
left middle finger:   --8---><---7--   :right ring finger        (BOTTOM)
  left ring finger:   --7--    --6--   :right little finger      (BOTTOM)
left little finger:   --6--                                      (BOTTOM)

Five bottom fingers make 5 tens, or 50. Two top left fingers and three top right fingers make the product 6. Summing these produces the answer, 56.
Another example, this time using 6 × 8:
 --8---><---6--
 --7--
 --6--

Four tens (bottom), plus two times four (top) gives 40 + 2 × 4 = 48.
Here's how it works: each finger represents a number between 6 and 10. When you join fingers representing x and y, there will be 10 - x "top" fingers and x - 5 "bottom" fingers on the left hand; the right hand will have 10 - y "top" fingers and y - 5 "bottom" fingers.
Let
t L = 10 − x {\displaystyle \,t_{L}=10-x\,} (the number of "top" fingers on the left hand)
t R = 10 − y {\displaystyle \,t_{R}=10-y\,} (the number of "top" fingers on the right hand)
b L = x − 5 {\displaystyle \,b_{L}=x-5\,} (the number of "bottom" fingers on the left hand)
b R = y − 5 {\displaystyle \,b_{R}=y-5\,} (the number of "bottom" fingers on the right hand)
Then following the above instructions produces
10 ( b L + b R ) + t L t R {\displaystyle \,10(b_{L}+b_{R})+t_{L}t_{R}}
= 10 [ ( x − 5 ) + ( y − 5 ) ] + ( 10 − x ) ( 10 − y ) {\displaystyle \,=10[(x-5)+(y-5)]+(10-x)(10-y)}
= 10 ( x + y − 10 ) + ( 100 − 10 x − 10 y + x y ) {\displaystyle \,=10(x+y-10)+(100-10x-10y+xy)}
= [ 10 ( x + y ) − 100 ] + [ 100 − 10 ( x + y ) + x y ] {\displaystyle \,=[10(x+y)-100]+[100-10(x+y)+xy]}
= [ 10 ( x + y ) − 10 ( x + y ) ] + [ 100 − 100 ] + x y {\displaystyle \,=[10(x+y)-10(x+y)]+[100-100]+xy}
= x y {\displaystyle \,=xy}
which is the product we seek.
Multiplying two numbers close and below 100[edit]
This technique allows easy multiplication of numbers close and below 100.(90-99)[3] The variables will be the two numbers you multiply.
The product of two variables ranging from 90-99 will result in a 4-digit number. The first step is to find the ones-digit and the tens digit.
Subtract both variables from 100 which will result in 2 one-digit number. The product of the 2 one-digit numbers will be the last two digits of your final product.
Next, subtract one of the two variables from 100. Then subtract the difference from the other variable. That difference will be the first two digits of your final product. And the resulting 4 digit number will be the final product.
Example:
          95
        x 97
        ----

Last two digits: 100-95=5 (subtract first number from 100)
                 100-97=3 (subtract second number from 100)
                 5*3=15   (multiply the two differences)
                 Final Product- yx15 First two digits: 100-95=5 (Subtract the first number of your equation from 100) 97-5=92 (Subtract that answer from the second number of your equation) Now, the difference will be the first two digits Final Product- 9215 Alternate for first two digits 5+3=8 (Add the two single digits derived when calculating "Last two digits" in previous step) 100-8=92 (Subtract that answer from 100) Now, the difference will be the first two digits Final Product- 9215

Using square numbers[edit]
The products of small numbers may be calculated by using the squares of integers; for example, to calculate 13 × 17, you can remark 15 is the mean of the two factors, and think of it as (15 − 2) × (15 + 2), i.e. 15² − 2². Knowing that 15² is 225 and 2² is 4, simple subtraction shows that 225 − 4 = 221, which is the desired product.
This method requires knowing by heart a certain number of squares:
12 = 1 62 = 36 112 = 121 162 = 256 212 = 441 262 = 676
22 = 4 72 = 49 122 = 144 172 = 289 222 = 484 272 = 729
32 = 9 82 = 64 132 = 169 182 = 324 232 = 529 282 = 784
42 = 16 92 = 81 142 = 196 192 = 361 242 = 576 292 = 841
52 = 25 102 = 100 152 = 225 202 = 400 252 = 625 302 = 900
Squaring numbers[edit]
It may be useful to be aware that the difference between two successive square numbers is the sum of their respective square roots. Hence if you know that 12 × 12 = 144 and wish to know 13 × 13, calculate 144 + 12 + 13 = 169.
This is because (x + 1)2 − x2 = x2 + 2x + 1 − x2 = x + (x + 1)
x2 = (x − 1)2 + (2x − 1)
Squaring any number[edit]
Take a given number, and add and subtract a certain value to it that will make it easier to multiply. For example:
4922
492 is close to 500, which is easy to multiply by. Add and subtract 8 (the difference between 500 and 492) to get
492 -> 484, 500
Multiply these numbers together to get 242,000 (This can be done efficiently by dividing 484 by 2 = 242 and multiplying by 1000). Finally, add the difference (8) squared (82 = 64) to the result:
4922 = 242,064
The proof follows:
n 2 = n 2 {\displaystyle n^{2}=n^{2}}
n 2 = ( n 2 − a 2 ) + a 2 {\displaystyle n^{2}=(n^{2}-a^{2})+a^{2}}
n 2 = ( n 2 − a n + a n − a 2 ) + a 2 {\displaystyle n^{2}=(n^{2}-an+an-a^{2})+a^{2}}
n 2 = ( n − a ) ( n + a ) + a 2 {\displaystyle n^{2}=(n-a)(n+a)+a^{2}}
Squaring any 2-digit integers[edit]
This method requires memorization of the squares of the one-digit numbers 1 to 9.
The square of mn, mn being a two-digit integer, can be calculated as
10 × m(mn + n) + n2
Meaning the square of mn can be found by adding n to mn, multiplied by m, adding 0 to the end and finally adding the square of n.
For example, we have 232:
232
= 10 × 2(23 + 3) + 32
= 10 × 2(26) + 9
= 520 + 9
= 529
So 232 = 529.
Squaring a number ending in 5[edit]
Take the digit(s) that precede the five: abc5, where a, b, and c are digits
Multiply this number by itself plus one: abc(abc + 1)
Take above result and attach 25 to the end
Example: 85 × 85
8
8 × 9 = 72
So, 852 = 7,225
Example: 1252
12
12 × 13 = 156
So, 1252 = 15,625
Mathematical explanation
(10x + 5)2 = (10x + 5)(10x + 5)
= 100x2 + 100x + 25
= 100(x2 + x) + 25
= 100x(x + 1) + 25
Squaring numbers near 50[edit]
Suppose we need to square a number x near 50. This number may be expressed as x = 50 − n, and hence the answer x2 is (50−n)2, which is 502 − 100n + n2. We know that 502 is 2500. So we subtract 100n from 2500, and then add n2. Example, say we want to square 48, which is 50 − 2. We subtract 200 from 2500 and add 4, and get x2 = 2304. For numbers larger than 50 (x = 50 + n), add n a hundred times instead of subtracting it.
Squaring an integer from 26 to 75[edit]
This method requires the memorization of squares from 1 to 25.
The square of n (most easily calculated when n is between 26 and 75 inclusive) is
(50 − n)2 + 100(n − 25)
In other words, the square of a number is the square of its difference from fifty added to one hundred times the difference of the number and twenty five. For example, to square 62, we have:
(−12)2 + [(62-25) × 100]
= 144 + 3,700
= 3,844
Squaring an integer from 76 to 125[edit]
This method requires the memorization of squares from 1 to 25.
The square of n (most easily calculated when n is between 76 and 125 inclusive) is
(100 − n)2 + 100(100 − 2(100 − n))
In other words, the square of a number is the square of its difference from one hundred added to the product of one hundred and the difference of one hundred and the product of two and the difference of one hundred and the number. For example, to square 93, we have:
72 + 100(100 − 2(7))
= 49 + 100 × 86
= 49 + 8,600
= 8,649
Another way to look at it would be like this:
932 = ? (is −7 from 100)
93 − 7 = 86 (this gives us our first two digits)
(−7)2 = 49 (these are the second two digits)
932 = 8649
Another example:
 822 = ? (is −18 from 100) 82 − 18 = 64 (subtract. First digits.) (−18)2 = 324 (second pair of digits. We'll need to carry the 3.) 82² = 6724

Finding roots[edit]
Approximating square roots[edit]
An easy way to approximate the square root of a number is to use the following equation:
root ≃ known square root − known square − unknown square 2 × known square root {\displaystyle {\text{root }}\simeq {\text{ known square root}}-{\frac {{\text{known square}}-{\text{unknown square}}}{2\times {\text{known square root}}}}\,}
The closer the known square is to the unknown, the more accurate the approximation. For instance, to estimate the square root of 15, we could start with the knowledge that the nearest perfect square is 16 (42).
root ≃ 4 − 16 − 15 2 × 4 ≃ 4 − 0.125 ≃ 3.875 {\displaystyle {\begin{aligned}{\text{root}}&\simeq 4-{\frac {16-15}{2\times 4}}\\&\simeq 4-0.125\\&\simeq 3.875\\\end{aligned}}\,\!}
So the estimated square root of 15 is 3.875. The actual square root of 15 is 3.872983... One thing to note is that, no matter what the original guess was, the estimated answer will always be larger than the actual answer due to the inequality of arithmetic and geometric means. Thus, one should try rounding the estimated answer down.
Note that if n2 is the closest perfect square to the desired square x and d = x - n2 is their difference, it is more convenient to express this approximation in the form of mixed fraction as n d 2 n {\displaystyle n{\tfrac {d}{2n}}} . Thus, in the previous example, the square root of 15 is 4 − 1 8 . {\displaystyle 4{\tfrac {-1}{8}}.} As another example, square root of 41 is 6 5 12 = 6.416 {\displaystyle 6{\tfrac {5}{12}}=6.416} while the actual value is 6.4031...
Derivation
By definition, if r is the square root of x, then
r 2 = x {\displaystyle \mathrm {r} ^{2}=x\,\!}
We then redefine the root
r = a − b {\displaystyle \mathrm {r} =a-b\,\!}
where a is a known root (4 from the above example) and b is the difference between the known root and the answer we seek.
( a − b ) 2 = x {\displaystyle (a-b)^{2}=x\,\!}
Expanding yields
a 2 − 2 a b + b 2 = x {\displaystyle a^{2}-2ab+b^{2}=x\,\!}
If 'a' is close to the target, 'b' will be a small enough number to render the + b 2 {\displaystyle {}+b^{2}\,} element of the equation negligible. Thus, we can drop + b 2 {\displaystyle {}+b^{2}\,} out and rearrange the equation to
b ≃ a 2 − x 2 a {\displaystyle b\simeq {\frac {a^{2}-x}{2a}}\,\!}
and therefore
r o o t ≃ a − a 2 − x 2 a {\displaystyle \mathrm {root} \simeq a-{\frac {a^{2}-x}{2a}}\,\!}
that can be reduced to
r o o t ≃ a 2 + x 2 a {\displaystyle \mathrm {root} \simeq {\frac {a^{2}+x}{2a}}\,\!}
Extracting roots of perfect powers[edit]
See also: 13th root
Extracting roots of perfect powers is often practiced. The difficulty of the task does not depend on the number of digits of the perfect power but on the precision, i.e. the number of digits of the root. In addition, it also depends on the order of the root; finding perfect roots, where the order of the root is coprime with 10 are somewhat easier since the digits are scrambled in consistent ways, as we shall see in the next section.
Extracting cube roots[edit]
An easy task for the beginner is extracting cube roots from the cubes of 2 digit numbers. For example, given 74088, determine what two digit number, when multiplied by itself once and then multiplied by the number again, yields 74088. One who knows the method will quickly know the answer is 42, as 423 = 74088.
Before learning the procedure, it is required that the performer memorize the cubes of the numbers 1-10:
13 = 1 23 = 8 33 = 27 43 = 64 53 = 125
63 = 216 73 = 343 83 = 512 93 = 729 103 = 1000
Observe that there is a pattern in the rightmost digit: adding and subtracting with 1 or 3. Starting from zero:
03 = 0
13 = 1 up 1
23 = 8 down 3
33 = 27 down 1
43 = 64 down 3
53 = 125 up 1
63 = 216 up 1
73 = 343 down 3
83 = 512 down 1
93 = 729 down 3
103 = 1000 up 1
There are two steps to extracting the cube root from the cube of a two digit number. Say you are asked to extract the cube root of 29791. Begin by determining the one's place (units) of the two digit number. You know it must be one, since the cube ends in 1, as seen above.
If perfect cube ends in 0, the cube root of it must end in 0.
If perfect cube ends in 1, the cube root of it must end in 1.
If perfect cube ends in 2, the cube root of it must end in 8.
If perfect cube ends in 3, the cube root of it must end in 7.
If perfect cube ends in 4, the cube root of it must end in 4.
If perfect cube ends in 5, the cube root of it must end in 5.
If perfect cube ends in 6, the cube root of it must end in 6.
If perfect cube ends in 7, the cube root of it must end in 3.
If perfect cube ends in 8, the cube root of it must end in 2.
If perfect cube ends in 9, the cube root of it must end in 9.
Note that every digit corresponds to itself except for 2, 3, 7 and 8, which are just subtracted from ten to obtain the corresponding digit.
The second step is to determine the first digit of the two digit cube root by looking at the magnitude of the given cube. To do this, remove the last three digits of the given cube (29791 → 29) and find the greatest cube it is greater than (this is where knowing the cubes of numbers 1-10 is needed). Here, 29 is greater than 1 cubed, greater than 2 cubed, greater than 3 cubed, but not greater than 4 cubed. The greatest cube it is greater than is 3, so the first digit of the two digit cube must be 3.
Therefore, the cube root of 29791 is 31.
Another example:
Find the cube root of 456533.
The cube root ends in 7.
After the last three digits are taken away, 456 remains.
456 is greater than all the cubes up to 7 cubed.
The first digit of the cube root is 7.
The cube root of 456533 is 77.
This process can be extended to find cube roots that are 3 digits long, by using arithmetic modulo 11.[4]
These types of tricks can be used in any root where the order of the root is coprime with 10; thus it fails to work in square root, since the power, 2, divides into 10. 3 does not divide 10, thus cube roots work.
Approximating common logs (log base 10)[edit]
To approximate a common log (to at least one decimal point accuracy), a few log rules, and the memorization of a few logs is required. One must know:
log(a x b) = log(a) + log(b)
log(a / b) = log(a) - log(b)
log(0) does not exist
log(1) = 0
log(2) ~ .30
log(3) ~ .48
log(7) ~ .85
From this information, one can find the log of any number 1-9.
log(1) = 0
log(2) ~ .30
log(3) ~ .48
log(4) = log(2 × 2) = log(2) + log(2) ~ .60
log(5) = log(10 / 2) = log(10) − log(2) ~ .70
log(6) = log(2 × 3) = log(2) + log(3) ~ .78
log(7) ~ .85
log(8) = log(2 × 2 × 2) = log(2) + log(2) + log(2) ~ .90
log(9) = log(3 × 3) = log(3) + log(3) ~ .96
log(10) = 1 + log(1) = 1
The first step in approximating the common log is to put the number given in scientific notation. For example, the number 45 in scientific notation is 4.5 x 10^1, but we will call it a x 10^b. Next, find the log of a, which is between 1 and 10. Start by finding the log of 4, which is .60, and then the log of 5, which is .70 because 4.5 is between these two. Next, and skill at this comes with practice, place a 5 on a logarithmic scale between .6 and .7, somewhere around .653 (NOTE: the actual value of the extra places will always be greater than if it were placed on a regular scale. i.e., you would expect it to go at .650 because it is halfway, but instead it will be a little larger, in this case .653) Once you have obtained the log of a, simply add b to it to get the approximation of the common log. In this case, a + b = .653 + 1 = 1.653. The actual value of log(45) ~ 1.65321.
The same process applies for numbers between 0 and 1. For example, 0.045 would be written as 4.5 × 10−2. The only difference is that b is now negative, so when adding you are really subtracting. This would yield the result 0.653 − 2, or −1.347.
Approximating natural exponents (log base e)[edit]
Natural exponents are used in many important expressions in modern science and engineering, with applications not limited to quantum mechanics, thermodynamics and signal communications. Using the laws of Natural exponents, memorization of the approximations below, and combination with other mental calculation methods, create a powerful and elegant means for changing complicated problems in the physical sciences into simple sums and products. The laws of Natural exponents (Exponentiation) are:
ea x eb = e (a+b) and e −a = 1/ ea and also
ea x e−b = e (a-b) = ea / eb
Table of approximations[edit]
Where possible, single digit numbers, followed by zeros, are used for ease of memorization, accuracy and to eliminate redundancy: (e4.5)2 = e9, e3 x e4 = e7, and e8 are used instead of (e4)2.
Exponent
Value
Error
Exponent
Value
Error
e3 ≈ 20 0.43%+ e23 ≈ 1010 0.11%+
e4 ≈ 55 0.74%− e25 ≈ 7.2x1010 0.0003%+
e4.25 ≈ 70 0.15%− e50.6 ≈ 1022 0.02%+
e4.5 ≈ 90 0.02%− e61 ≈ π x 1026 0.02%−
e5.75 ≈ 100π 0.01%− e99 ≈ 1043 0.01%−
e8 ≈ 3,000 0.64%− e108 ≈ 8x1046 0.0015%−
Optimized Relations[edit]
This table presents optimized suggestions derived from the above table.
e10 = e3x e7 ≈ 22,000 0.01%+ e0.25 = e4.25 /e4 ≈ 70/55 3.5%+
e1 = e10 /e9 ≈ 220/81 0.08%+ e0.5 = e4.5 /e4 ≈ 90/55 1.5%+
e2 = e9 / e7 ≈ 81/11 0.4%− e0.3 = √e50.6 /e25 = e25.3 /e25 ≈ 100/72 0.5%+
e5 = e9 /e4 ≈ 8100/55 0.15%− better approximations: e4 ≈ 54.6 e6 ≈ 403 or 404
Physical Science and Communication Approximations[edit]
Exponent
Value
Error
Application
eπ ≈ 20+π 0.004%− Spherical wave attenuation
e274 ≈ 10119 0.0003%+ 1 °C, 274 Kelvin
e280 ≈ 4 x 10121 0.00003%+ e280/e7 = e273 near ambient, Kelvin
e452 ≈ 2 x 10196 0.000004%+ Accurate thermal temperature
kbe2 ≈ 10−22 1.4%+ kb = Boltzmann's constant
The +/- superscript after the error represents if the actual number value is higher or lower than the approximation; for example the approximation of ln(400) is less than 6. The +/- symbols after errors may also be used to make the approximations more accurate by compensation, for example;
e16 = e8 x e8, accurate to 12 parts in 1000, may be made more accurate by using
e7 x e9 ≈ 1,100 x 8,100 or 8,910,000, accuracy 2 parts in 1000.
Other examples[edit]
e(9 + π) = 8,100 x (20 + π) = 162,000 + ~25,500 = 187,500, (accurate to 1 part in 5000)
Mental arithmetic as a psychological skill[edit]
Physical exertion of the proper level can lead to an increase in performance of a mental task, like doing mental calculations, performed afterward.[5] It has been shown that during high levels of physical activity there is a negative effect on mental task performance.[6] This means that too much physical work can decrease accuracy and output of mental math calculations. Physiological measures, specifically EEG, have been shown to be useful in indicating mental workload.[7] Using an EEG as a measure of mental workload after different levels of physical activity can help determine the level of physical exertion that will be the most beneficial to mental performance. Previous work done at Michigan Technological University by Ranjana Mehta includes a recent study that involved participants engaging in concurrent mental and physical tasks.[8] This study investigated the effects of mental demands on physical performance at different levels of physical exertion and ultimately found a decrease in physical performance when mental tasks were completed concurrently, with a more significant effect at the higher level of physical workload. The Brown-Peterson procedure is a widely known task using mental arithmetic. This procedure, mostly used in cognitive experiments, suggests mental subtraction is useful in testing the effects maintenance rehearsal can have on how long short-term memory lasts.
Mental Calculations World Championship[edit]
The first Mental Calculations World Championship took place in 1997. This event repeats every year. It consists of a range of different tasks such as: addition of ten ten-digit numbers, multiplication of two eight-digit numbers, calculation of square roots, and calculation of weekdays for given dates, calculation of cube roots plus some surprise miscellaneous tasks.
Mental Calculation World Cup[edit]
Main article: Mental Calculation World Cup
The first World Mental Calculation Championships (Mental Calculation World Cup) [9] took place in 2004. They are repeated every second year. It consists of six different tasks: addition of ten ten-digit numbers, multiplication of two eight-digit numbers, calculation of square roots, and calculation of weekdays for given dates, calculation of cube roots plus some surprise miscellaneous tasks.
Memoriad – World Memory, Mental Calculation & Speed Reading Olympics[edit]
Memoriad[10] is the first platform combining "mental calculation", "memory" and "photographic reading" competitions. Games and competitions are held in the year of the Olympic games, every four years. The first Memoriad was held in Istanbul, Turkey, in 2008. The second Memoriad took place in Antalya, Turkey on 24–25 November 2012. 89 competitors from 20 countries participated. Awards and money prizes were given for 10 categories in total; of which 5 categories had to do about Mental Calculation (Mental addition, Mental Multiplication, Mental Square Roots (non-integer), Mental Calendar Dates calculation and Flash Anzan).
See also[edit]
Doomsday rule for calculating the day of the week
Mental abacus
Mental calculator
Soroban
References[edit]
^ Jump up to: a b c Cheprasov, Artem (September 3, 2009). On a New Method of Multiplication and Shortcuts. United States: CreateSpace Independent Publishing Platform. ISBN 9781448689330.
Jump up ^ "On the record with ... Artem Cheprasov". Northwest Herald. Retrieved 2015-06-01.
Jump up ^ multiplying two numbers close, below 100
Jump up ^ Dorrell, Philip. "How to Do Cube Roots of 9 Digit Numbers in Your Head". Thinking Hard. Retrieved 19 July 2015.
Jump up ^ Lambourne, Kate; Tomporowski, Phillip (2010). "The effect of exercise-induced arousal on cognitive task performance: A meta-regression analysis". Brain Research. 1341: 12–24. doi:10.1016/j.brainres.2010.03.091. PMID 20381468.
Jump up ^ Brisswalter, J.; Arcelin, R.; Audiffren, M.; Delignieres, D. (1997). "Influence of Physical Exercise on Simple Reaction Time: Effect of Physical Fitness". Perceptual and Motor Skills. 85 (3): 1019–27. doi:10.2466/pms.1997.85.3.1019. PMID 9399313.
Jump up ^ Murata, Atsuo (2005). "An Attempt to Evaluate Mental Workload Using Wavelet Transform of EEG". Human Factors: the Journal of the Human Factors and Ergonomics Society. 47 (3): 498. doi:10.1518/001872005774860096.
Jump up ^ Mehta, Ranjana K.; Nussbaum, Maury A.; Agnew, Michael J. (2012). "Muscle- and task-dependent responses to concurrent physical and mental workload during intermittent static work". Ergonomics. 55 (10): 1166–79. doi:10.1080/00140139.2012.703695. PMID 22849301.
Jump up ^ Mental Calculation World Cup
Jump up ^ Memoriad
External links[edit]
Mental Calculation World Cup
Memoriad - World Mental Olympics
World Mental Calculation directory
Tzourio-Mazoyer, Nathalie; Pesenti, Mauro; Zago, Laure; Crivello, Fabrice; Mellet, Emmanuel; Samson, Dana; Duroux, Bruno; Seron, Xavier; Mazoyer, Bernard (2001). "Mental calculation in a prodigy is sustained by right prefrontal and medial temporal areas". Nature Neuroscience. 4 (1): 103–7. doi:10.1038/82831. PMID 11135652.
Rivera, S.M.; Reiss, AL; Eckert, MA; Menon, V (2005). "Developmental Changes in Mental Arithmetic: Evidence for Increased Functional Specialization in the Left Inferior Parietal Cortex". Cerebral Cortex. 15 (11): 1779–90. doi:10.1093/cercor/bhi055. PMID 15716474.
Large EEG waves ellicited by Mental Calculation PDF
Mathletics - train or compete in Mental Math
Mathematical Shortcuts from Vedic Maths
Authority control
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NDL: 00560277
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Challenger Deep
From Wikipedia, the free encyclopedia
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Location of Challenger Deep within the Mariana Trench and Western Pacific Ocean
The Challenger Deep is the deepest known point in the Earth's seabed hydrosphere, with a depth of 10,898 to 10,916 m (35,755 to 35,814 ft) by direct measurement from submersibles, and slightly more by sonar bathymetry. It is in the Pacific Ocean, at the southern end of the Mariana Trench near the Mariana Islands group. The Challenger Deep is a relatively small slot-shaped depression in the bottom of a considerably larger crescent-shaped oceanic trench, which itself is an unusually deep feature in the ocean floor. Its bottom is about 11 km (7 mi) long and 1.6 km (1 mi) wide, with gently sloping sides.[1] The closest land to the Challenger Deep is Fais Island (one of the outer islands of Yap), 287 km (178 mi) southwest, and Guam, 304 km (189 mi) to the northeast. It is located in the ocean territory of the Federated States of Micronesia, 1.6 km (1 mi) from its border with ocean territory associated with Guam.[2]
The depression is named after the British Royal Navy survey ship HMS Challenger, whose expedition of 1872–1876 made the first recordings of its depth. According to the August 2011 version of the GEBCO Gazetteer of Undersea Feature Names, the location and depth of the Challenger Deep are 11°22.4′N 142°35.5′E / 11.3733°N 142.5917°E / 11.3733; 142.5917 and 10,920 m (35,827 ft) ±10 m (33 ft).[3]
June 2009 sonar mapping of the Challenger Deep by the Simrad EM120 (sonar multibeam bathymetry system for 300–11,000 m deep water mapping) aboard the RV Kilo Moana indicated a depth of 10,971 metres (35,994 ft). The sonar system uses phase and amplitude bottom detection, with a precision of 0.2% to 0.5% of water depth; this is an error of about 22 to 55 m (72 to 180 ft) at this depth.[4][5][6] Further soundings made by the US Center for Coastal & Ocean Mapping in October 2010 are in agreement with this figure, preliminarily placing the deepest part of the Challenger Deep at 10,994 m (36,070 ft), with an estimated vertical uncertainty of ±40 m (131 ft).[7][8] A 2014 study concludes that with the best of 2010 multibeam echosounder technologies a depth uncertainty of ±25 m (82 ft) (95% confidence level) on 9 degrees of freedom and a positional uncertainty of ±20 to 25 m (66 to 82 ft) (2drms) remain and the location of the deepest depth recorded in the 2010 mapping is 10,984 m (36,037 ft) at 11°19′48″N 142°11′57″E / 11.329903°N 142.199305°E / 11.329903; 142.199305 (11°19′47.650″N 142°11′57.498″E / 11.32990278°N 142.19930500°E / 11.32990278; 142.19930500).[6]
Only four descents have ever been achieved. The first descent by any vehicle was by the manned bathyscaphe Trieste in 1960. This was followed by the unmanned ROVs Kaikō in 1995 and Nereus in 2009. In March 2012 a manned solo descent was made by film director James Cameron in the deep-submergence vehicle Deepsea Challenger.[9][10][11] These expeditions measured very similar depths of 10,898 to 10,916 metres (35,755 to 35,814 ft).
Contents
1 History of depth mapping from the surface
2 Descents
2.1 Manned descents
2.1.1 Trieste
2.1.2 Deepsea Challenger
2.1.3 Planned manned descents
2.2 Unmanned descents
2.2.1 Kaikō
2.2.2 ABISMO
2.2.3 Nereus
2.2.4 Haidou-1
3 Lifeforms
4 See also
5 Notes
6 External links
History of depth mapping from the surface[edit]
Over many years, the search for the point of maximum depth has involved many different vessels.[12]
The Challenger expedition (December 1872 – May 1876) first sounded the depths now known as the Challenger Deep. This first sounding was made on 23 March 1875 at station 225.[13] The reported depth was 4,475 fathoms (26,850 ft; 8,184 m) at 11°24′N 143°16′E / 11.400°N 143.267°E / 11.400; 143.267, based on two separate soundings.
A 1912 book, The Depths of the Ocean by Sir John Murray, records the depth of the Challenger Deep as 31,614 ft (9,636 m), reporting the sounding taken by the converted navy collier USS Nero in 1899.[14] Murray was one of the expedition scientists.[15]
In 1951, about 75 years after its original discovery, the entire Mariana Trench was surveyed by a second Royal Navy vessel, captained by George Stephen Ritchie (later Rear Admiral Ritchie); this vessel was also named HMS Challenger, after the original expedition ship. This survey recorded the deepest part of the trench using echo sounding, a more precise and easier way to measure depth than the sounding equipment and drag lines used in the original expedition. A depth of 5,960 fathoms (35,760 ft; 10,900 m) was measured at 11°19′N 142°15′E / 11.317°N 142.250°E / 11.317; 142.250.
Research vessel Vityaz in Kaliningrad "Museum of world ocean"
The maximum surveyed depth of the Challenger Deep was reported in 1957 by the Soviet Research vessel Vityaz recording a spot 11,034 metres (36,201 ft) ±50 m (164 ft) deep at 11°20.9′N 142°11.5′E / 11.3483°N 142.1917°E / 11.3483; 142.1917.[12] It was dubbed the Mariana Hollow and is listed in many reference sources, including the Encyclopædia Britannica,[16] articles in National Geographic[17] and on maps. The pressure at this depth is approximately 1,099 times that at the surface, or 111 MPa (16,099 psi).[18]
In 1959, the US Navy research vessel RV Stranger using bomb-sounding surveyed a maximum depth of 10,915 m (35,810 ft) ±10 m (33 ft) at 11°20.0′N 142°11.8′E / 11.3333°N 142.1967°E / 11.3333; 142.1967.[3][12]
In 1962, the US Navy research vessel RV Spencer F. Baird using a frequency-controlled depth recorder surveyed a maximum depth of 10,915 m (35,810 ft) ±10 m (33 ft) at 11°20.0′N 142°11.8′E / 11.3333°N 142.1967°E / 11.3333; 142.1967.[12]
In 1975 and 1980, the US Navy research vessel RV Thomas Washington using a precision depth recorder with satellite positioning surveyed a maximum depth of 10,915 m (35,810 ft) ±10 m (33 ft) at 11°20.0′N 142°11.8′E / 11.3333°N 142.1967°E / 11.3333; 142.1967.[12]
In 1984, the survey vessel Takuyo from the Hydrographic Department of Japan, used a narrow, multibeam echo sounder to take a measurement of 10,924 m (35,840 ft) ±10 m (33 ft) at 11°22.4′N 142°35.5′E / 11.3733°N 142.5917°E / 11.3733; 142.5917.[3][12][19]
Deep Sea Research Vessel RV Kairei
In 1998, a regional bathymetric survey of the Challenger Deep was conducted by the Deep Sea Research Vessel RV Kairei, from the Japan Agency for Marine-Earth Science and Technology, using a SeaBeam 2112 multibeam echosounder. The regional bathymetric map made from the data obtained in 1998 shows that the greatest depths in the eastern, central, and western depressions are 10,922 m (35,833 ft) ±74 m (243 ft), 10,898 m (35,755 ft) ±62 m (203 ft), and 10,908 m (35,787 ft) ±36 m (118 ft), respectively, making the eastern depression the deepest of the three.[12]
In 1999 and 2002, Kairei revisited the Challenger Deep. The cross track survey in the 1999 Kairei cruise shows that the greatest depths in the eastern, central, and western depressions are 10,920 m (35,827 ft) ±10 m (33 ft), 10,894 m (35,741 ft) ±14 m (46 ft), and 10,907 m (35,784 ft) ±13 m (43 ft), respectively, which supports the results of the 1998 survey. The detailed grid survey in 2002 showed that the deepest site is located in the eastern part of the eastern depression around 11°22.260′N 142°35.589′E / 11.371000°N 142.593150°E / 11.371000; 142.593150, with a depth of 10,920 m (35,827 ft) ±5 m (16 ft), about 290 m (950 ft) southeast of the deepest site determined by the survey vessel Takuyo in 1984 and about 240 m (790 ft) east of the deepest place determined by the 1998 Kairei survey.[12]
On 1 June 2009, sonar mapping of the Challenger Deep by the Kongsberg Simrad EM 120 sonar multibeam bathymetry system for deep water (300 – 11,000 metres) mapping aboard RV Kilo Moana (mothership of the underwater vehicle Nereus) indicated a depth of 10,971 m (35,994 ft). The sonar system uses phase and amplitude bottom detection, which is capable of an accuracy of 0.2% to 0.5% of water depth across the entire swath.[4][5][20] In 2014 the multibeam bathymetry data of this sonar mapping have yet to be publicly released, so the data are not available for comparisons with other soundings.[6]
On 7 October 2010, further sonar mapping of the Challenger Deep area was conducted by the US Center for Coastal & Ocean Mapping/Joint Hydrographic Center (CCOM/JHC) aboard USNS Sumner. The results were reported in December 2011 at the annual American Geophysical Union fall meeting. Using a Kongsberg Maritime EM 122 multibeam echosounder system coupled to positioning equipment that can determine latitude and longitude up to 50 cm (20 in) accuracy, from thousands of individual soundings around the deepest part the CCOM/JHC team preliminary determined that the Challenger Deep has a maximum depth of 10,994 m (36,070 ft) at 11°19′35″N 142°11′14″E / 11.326344°N 142.187248°E / 11.326344; 142.187248, with an estimated vertical uncertainty of ±40 m (131 ft) at 2 standard deviations (≈ 95.4%) confidence level.[7] A secondary deep with a depth of 10,951 m (35,928 ft) was located at approximately 23.75 nmi (44.0 km) to the east at 11°22′11″N 142°35′19″E / 11.369639°N 142.588582°E / 11.369639; 142.588582 in the Mariana Trench.[8][21][22][23]
In 2014, a study was conducted regarding the determination of the depth and location of the Challenger Deep based on data collected previous to and during the 2010 sonar mapping of the Mariana Trench with a Kongsberg Maritime EM 122 multibeam echosounder system aboard USNS Sumner. This study by James. V. Gardner et al. of the Center for Coastal & Ocean Mapping-Joint Hydrographic Center (CCOM/JHC), Chase Ocean Engineering Laboratory of the University of New Hampshire splits the measurement attempt history into three main groups: early single-beam echo sounders (1950s-1970s), early multibeam echo sounders (1980s - 21st century), and modern (i.e., post-GPS, high-resolution) multibeam echo sounders. Taking uncertainties in depth measurements and position estimation into account the raw data of the 2010 bathymetry of the Challenger Deep vicinity consisting of 2,051,371 soundings from eight survey lines was analyzed. The study concludes that with the best of 2010 multibeam echosounder technologies after the analysis a depth uncertainty of ±25 m (82 ft) (95% confidence level) on 9 degrees of freedom and a positional uncertainty of ±20 to 25 m (66 to 82 ft) (2drms) remain and the location of the deepest depth recorded in the 2010 mapping is 10,984 m (36,037 ft) at 11°19′48″N 142°11′57″E / 11.329903°N 142.199305°E / 11.329903; 142.199305. The depth measurement uncertainty is a composite of measured uncertainties in the spatial variations in sound-speed through the water volume, the ray-tracing and bottom-detection algorithms of the multibeam system, the accuracies and calibration of the motion sensor and navigation systems, estimates of spherical spreading, attenuation throughout the water volume, and so forth.[6]
The 2009 and 2010 maximal depths were not confirmed by the series of dives Nereus made to the bottom during an expedition in May–June 2009. The direct descent measurements by the four expeditions which have reported from the bottom, have fixed depths in a narrow range from 10,916 m (Trieste) to 10,911 m (Kaikō), to 10,902 m (Nereus) to 10,898 m (Deepsea Challenger) Although an attempt was made to correlate locations, it could not be absolutely certain that Nereus (or the other descents) reached exactly the same points found to be maximally deep by the sonar/echo sounders of previous mapping expeditions, even though one of these echo soundings was made by Nereus' mothership.
Descents[edit]
Manned descents[edit]
Lt. Don Walsh, USN (bottom) and Jacques Piccard (center) in the bathyscaphe Trieste.
Trieste[edit]
Main article: Bathyscaphe Trieste
On 23 January 1960, the Swiss-designed Trieste, originally built in Italy and acquired by the U.S. Navy, descended to the ocean floor in the trench manned by Jacques Piccard (who co-designed the submersible along with his father, Auguste Piccard) and USN Lieutenant Don Walsh. Their crew compartment was inside a spherical pressure vessel, which was a heavy-duty replacement (of the Italian original) built by Krupp Steel Works of Essen, Germany. Their descent took almost five hours and the two men spent barely twenty minutes on the ocean floor before undertaking the three-hour-and-fifteen-minute ascent. Their early departure from the ocean floor was due to their concern over a crack in the outer window caused by the temperature differences during their descent.[24] The measured depth at the bottom was measured with a manometer at 10,916 m (35,814 ft) ±5 m (16 ft).[12][25]
Deepsea Challenger[edit]
Main article: Deepsea Challenger
DSV Deepsea Challenger
On 26 March 2012 (local time), Canadian film director James Cameron made a solo manned descent in the DSV Deepsea Challenger to the bottom of the Challenger Deep.[9][10][11][26] At approximately 05:15 ChST on 26 March (19:15 UTC on 25 March), the descent began.[27] At 07:52 ChST (21:52 UTC), Deepsea Challenger arrived at the bottom. The descent lasted 2 hours and 36 minutes and the recorded depth was 10,898.4 metres (35,756 ft) when Deepsea Challenger touched down.[28] Cameron had planned to spend about six hours near the ocean floor exploring but decided to start the ascent to the surface after only 2 hours and 34 minutes.[29] The time on the bottom was shortened because a hydraulic fluid leak in the lines controlling the manipulator arm obscured the visibility out the only viewing port. It also caused the loss of the submersible's starboard thrusters.[30] At around 12:00 ChST (02:00 UTC on 26 March), the Deepsea Challenger website says the sub resurfaced after a 90-minute ascent,[31] although Paul Allen's tweets indicate the ascent took only about 67 minutes.[32] During a post-dive press conference Cameron said: "I landed on a very soft, almost gelatinous flat plain. Once I got my bearings, I drove across it for quite a distance ... and finally worked my way up the slope." The whole time, Cameron said, he didn't see any fish, or any living creatures more than an inch (2.54 cm) long: "The only free swimmers I saw were small amphipods"—shrimplike bottom-feeders.[33]
Planned manned descents[edit]
Several other manned expeditions are planned. These include:[34]
Triton Submarines, a Florida-based company that designs and manufactures private submarines, whose vehicle, Triton 36000/3, will carry a crew of three to the seabed in 120 minutes;[35]
Virgin Oceanic, sponsored by Richard Branson's Virgin Group, is developing a submersible designed by Graham Hawkes, DeepFlight Challenger,[36] with which the solo pilot will take 140 minutes to reach the seabed;[37]
DOER Marine, a San Francisco Bay Area based marine technology company established in 1992, that is developing a vehicle, Deepsearch (and Ocean Explorer HOV Unlimited), with some support from Google's Eric Schmidt with which a crew of two or three will take 90 minutes to reach the seabed, as the program Deep Search.[38]
Unmanned descents[edit]
Kaikō[edit]
Main article: Kaikō
On 24 March 1995, the Japanese robotic deep-sea probe Kaikō broke the depth record for unmanned probes when it reached close to the surveyed bottom of the Challenger Deep. Created by the Japan Agency for Marine-Earth Science and Technology (JAMSTEC), it was one of the few unmanned deep-sea probes in operation that could dive deeper than 6,000 metres (20,000 ft). The manometer measured depth of 10,911 m (35,797 ft) ±3 m (10 ft) at 11°22.39′N 142°35.54′E / 11.37317°N 142.59233°E / 11.37317; 142.59233[39] for the Challenger Deep is believed to be the most accurate measurement taken yet.[12] Kaikō also collected sediment cores containing marine organisms from the bottom of the deep.[17][18] Kaikō made many unmanned descents to the Mariana Trench during three expeditions in 1995, 1996 and 1998.[40] The greatest depth measured by Kaikō in 1996 was 10,898 m (35,755 ft) at 11°22.10′N 142°25.85′E / 11.36833°N 142.43083°E / 11.36833; 142.43083 and in 1998 10,907 m (35,784 ft) at 11°22.95′N 142°12.42′E / 11.38250°N 142.20700°E / 11.38250; 142.20700.[12] It was lost at sea off Shikoku Island during Typhoon Chan-Hom on 29 May 2003.
ABISMO[edit]
Main article: ABISMO
On 3 June 2008, the Japanese robotic deep-sea probe ABISMO (Automatic Bottom Inspection and Sampling Mobile) reached the bottom of the Mariana Trench about 150 km (93 mi) east of the Challenger Deep and collected core samples of the deep sea sediment and water samples of the water column. Created by the Japan Agency for Marine-Earth Science and Technology (JAMSTEC), it was the only unmanned deep-sea probe in use that could dive deeper than 10,000 m (32,808 ft) after that of Nereus. During ABISMO's deepest Mariana Trench dive its manometer measured a depth of 10,258 m (33,655 ft) ±3 m (10 ft)[41]
Nereus[edit]
HROV Nereus
Main article: Nereus (underwater vehicle)
On 31 May 2009 the United States sent the Nereus hybrid remotely operated vehicle (HROV) to the Challenger Deep.[42] Nereus thus became the first vehicle to reach the Mariana Trench since 1998 and the deepest-diving vehicle then in operation.[42] Project manager and developer Andy Bowen heralded the achievement as "the start of a new era in ocean exploration".[42] Nereus, unlike Kaikō, did not need to be powered or controlled by a cable connected to a ship on the ocean surface.[43]
Nereus spent over 10 hours at the bottom of the Challenger Deep and measured a depth of 10,902 m (35,768 ft) at 11°22.1′N 142°35.4′E / 11.3683°N 142.5900°E / 11.3683; 142.5900, while sending live video and data back to its mothership RV Kilo Moana at the surface and collecting geological and biological samples from the Challenger Deep bottom with its manipulator arm for further scientific analysis.[4][42][44][45]
The Nereus was operated by the Woods Hole Oceanographic Institution. It was lost on May 10, 2014.[46]
Haidou-1[edit]
On 23 May 2016 the Chinese submersible Haidou-1 dove to a depth of 10,767 m (35,325 ft) in the Mariana Trench. This autonomous and remotely operated vehicle has a design depth of 11,000 m (36,089 ft).[47]
Lifeforms[edit]
The Summary Report of the HMS Challenger expedition lists radiolaria from the two dredged samples taken when the Challenger Deep was first discovered.[48] These (Nassellaria and Spumellaria) were reported in the Report on Radiolaria (1887)[49] written by Ernst Haeckel.
On their 1960 descent, the crew of the Trieste noted that the floor consisted of diatomaceous ooze and reported observing "some type of flatfish" lying on the seabed.[50]
"... And as we were settling this final fathom, I saw a wonderful thing. Lying on the bottom just beneath us was some type of flatfish, resembling a sole, about 1 foot long and 6 inches across. Even as I saw him, his two round eyes on top of his head spied us — a monster of steel — invading his silent realm. Eyes? Why should he have eyes? Merely to see phosphorescence? The floodlight that bathed him was the first real light ever to enter this hadal realm. Here, in an instant, was the answer that biologists had asked for the decades. Could life exist in the greatest depths of the ocean? It could! And not only that, here apparently, was a true, bony teleost fish, not a primitive ray or elasmobranch. Yes, a highly evolved vertebrate, in time's arrow very close to man himself. Slowly, extremely slowly, this flatfish swam away. Moving along the bottom, partly in the ooze and partly in the water, he disappeared into his night. Slowly too — perhaps everything is slow at the bottom of the sea — Walsh and I shook hands.[51]
Many marine biologists are now skeptical of this supposed sighting, and it is suggested that the creature may instead have been a sea cucumber.[52][53] The video camera on board the Kaiko probe spotted a sea cucumber, a scale worm and a shrimp at the bottom.[54][55] At the bottom of the Challenger deep, the Nereus probe spotted one polychaete worm (a multi-legged predator) about an inch long.[56]
An analysis of the sediment samples collected by Kaiko found large numbers of simple organisms at 10,900 m (35,800 ft).[57] While similar lifeforms have been known to exist in shallower ocean trenches (> 7,000 m) and on the abyssal plain, the lifeforms discovered in the Challenger Deep possibly represent taxa distinct from those in shallower ecosystems.
Most of the organisms collected were simple, soft-shelled foraminifera (432 species according to National Geographic[58]), with four of the others representing species of the complex, multi-chambered genera Leptohalysis and Reophax. Eighty-five percent of the specimens were organic, soft-shelled allogromiids, which is unusual compared to samples of sediment-dwelling organisms from other deep-sea environments, where the percentage of organic-walled foraminifera ranges from 5% to 20%. As small organisms with hard, calcareous shells have trouble growing at extreme depths because of the high solubility of calcium carbonate in the pressurized water, scientists theorize that the preponderance of soft-shelled organisms in the Challenger Deep may have resulted from the typical biosphere present when the Challenger Deep was shallower than it is now. Over the course of six to nine million years, as the Challenger Deep grew to its present depth, many of the species present in the sediment died out or were unable to adapt to the increasing water pressure and changing environment.[59] The species that survived the change in depth were the ancestors of the Challenger Deep's current denizens.[citation needed]
On 17 March 2013, researchers reported data that suggested microbial life forms thrive in the Challenger Deep.[60][61] Other researchers reported related studies that microbes thrive inside rocks up to 1,900 ft (579 m) below the sea floor under 8,500 ft (2,591 m) of ocean off the coast of the northwestern United States.[60][62] According to one of the researchers, "You can find microbes everywhere — they're extremely adaptable to conditions, and survive wherever they are."[60]
See also[edit]
Sirena Deep
Horizon Deep
Galathea Depth
Notes[edit]
Jump up ^ All Things Considered (9 March 2012). "Diving Back to the Bottom of the Mariana Trench". NPR. Retrieved 26 March 2012.
Jump up ^ The Colbert Report, airdate: 2012 April 12, interview with James Cameron
^ Jump up to: a b c "IHO-IOC GEBCO Gazetteer of Undersea Feature Names, August 2011 version". GEBCO. August 2011. Retrieved 20 March 2012.
^ Jump up to: a b c "Daily Reports for R/V KILO MOANA June and July 2009". University of Hawaii Marine Center. 4 June 2009. Retrieved 4 June 2009.
^ Jump up to: a b "Inventory of Scientific Equipment aboard the R/V KILO MOANA". University of Hawaii Marine Center. 4 June 2009. Retrieved 4 June 2009.
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^ Jump up to: a b Amos, Jonathan (7 December 2011). "Oceans' deepest depth re-measured". BBC News. Retrieved 7 December 2011.
^ Jump up to: a b Armstrong, Andrew A. (2011-12-22). "Cruise Report - UNH-CCOM/JHC Technical Report 11-002" (PDF). NOAA/UNH Joint Hydrographic Center University of New Hampshire. p. 12. Retrieved 2012-05-01.
^ Jump up to: a b Than, Ker (25 March 2012). "James Cameron Completes Record-Breaking Mariana Trench Dive". National Geographic Society. Retrieved 25 March 2012.
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^ Jump up to: a b c Choi, Charles Q. (17 March 2013). "Microbes Thrive in Deepest Spot on Earth". LiveScience. Retrieved 17 March 2013.
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External links[edit]
"Official press release regarding Challenger Deep operation". Archived from the original on 18 April 2002.
Mariana Trench
Mariana Trench: Seven miles deep, the ocean is still a noisy place
Location of Challenger Deep Interactive map
Coordinates: 11°22.4′N 142°35.5′E / 11.3733°N 142.5917°E / 11.3733; 142.5917
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