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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 other supplies, for example, a calculator. People use mental calculation when computing tools are not available, when it is faster than other means of calculation (for example, conventional methods as taught in educational institutions), or 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 determines what methods to use and also which calculations are easier to perform mentally.In this case you can not use any supplies such as calculator if you do it will no make you any better at learning. In the future if you didn't do it your selve it can cause a lot of problems. For your learning
Contents
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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.
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 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.

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.
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. PMID 20381468. doi:10.1016/j.brainres.2010.03.091.
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. PMID 9399313. doi:10.2466/pms.1997.85.3.1019.
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. PMID 22849301. doi:10.1080/00140139.2012.703695.
Jump up ^ Mental Calculation World Cup
Jump up ^ Memoriad
External links[edit]
Mental Calculation World Cup
Memoriad - World Mental Olympics
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. PMID 11135652. doi:10.1038/82831.
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. PMID 15716474. doi:10.1093/cercor/bhi055.
Large EEG waves ellicited by Mental Calculation PDF
Mathletics - train or compete in Mental Math
Mathematical Shortcuts from Vedic Maths
Authority control
GND: 4165264-2
NDL: 00560277
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Two Out of Three Ain't Bad
From Wikipedia, the free encyclopedia
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This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed. (October 2013) (Learn how and when to remove this template message)
"Two Out of Three Ain't Bad"
Side-A label of U.S. 7-inch vinyl single
Single by Meat Loaf
from the album Bat Out of Hell
B-side
"For Crying Out Loud"
Released
February 1978
Format
7-inch single
Length
5:27 (LP version)
3:58 (7" version)
Label
Epic
Songwriter(s)
Jim Steinman
Producer(s)
Todd Rundgren
Meat Loaf singles chronology
"You Took the Words Right Out of My Mouth"
(1977) "Two Out of Three Ain't Bad"
(1978) "Paradise by the Dashboard Light"
(1978)
"You Took the Words Right Out of My Mouth"
(1977) "Two Out of Three Ain't Bad"
(1978) "Paradise by the Dashboard Light"
(1978)
"Two Out of Three Ain't Bad" is a love song performed by the American musician Meat Loaf. It is a track off his 1977 album Bat Out of Hell, written by Jim Steinman. It reached #11 on the Billboard Hot 100 and earned a million-selling Gold single from the RIAA. It stands as one of his career signature tunes, still enjoying recurrent airplay.
It was the final song written for the album. In a 2003 interview for the VH1 Ultimate Albums series, Steinman recalls:
"I remember Mimi Kennedy [a cast member of Jim's then-current musical Rhinegold] telling me, she said, you know, when I was probably complaining why no one liked my stuff and couldn't get a deal, she says, "Well Steiny, your stuff is so complicated. Can't you write something simple?" And while she was saying that the oldies station was on the radio and it was playing that old Elvis song, 'I Want You, I Need,' whatever it was. 'I Want You, I Need You, I Love You', you know. I just started singing my own song but it was 'I Want You, I Need You, I Love You.' She said, "Why don't you write something simple like that, 'I want you, I need you, I love you'?" I said, "Well I'll try." I don't try to make them complicated. I remember going home and I tried so hard but the best I could do was: I want you, I need you but there ain't no way I'm ever gonna love you, don't be sad, 'cause two out of three ain't bad. So it was still a twist but it was my closest to a simple song, and one Elvis could have done."
A music video was shot for the single and received significant airplay on television video programs of that day, a number of years before the conception of MTV.
Two mixes were created for the song. In the album mix, timed around 5:27, one can hear prominent Moog-synthesizer embellishments by session keyboardist Roger Powell. The single mix, timed around 3:50, downmixes the synthesizer considerably, in place of a much warmer and predominant string arrangement by Ken Ascher. This mix also edits out and removes the first lyrical bridge of the song which includes the line, "There ain't no Coupe de Ville hiding at the bottom of a Cracker Jack box", and does some overlapping, splicing and editing to seamlessly merge back into the second part of the lyrical bridge, starting with, "I can't lie, I can't tell you that I'm something I'm not". Yet, another lyrical edit in the last verse removes the lines, "And though I know I'll never get her out of my heart, she never loved me back, Ooh I know, well I remember how she left me on a stormy night, oh she kissed me and got out of our bed". Lastly, a song fade-out beginning in the middle of the ending chorus' second refrain is also featured, hence removing the full-circle lyrics of "Baby we can talk all night, but that ain't getting us nowhere," that both started and ended the song, in its full album mix version.
The Japanese release changes the title of the song to use a percentage, translating it as "66% is good enough".[1]
Contents
1 Subsequent versions
2 Charts
2.1 Meat Loaf version
2.1.1 Weekly charts
2.1.2 Year-end charts
2.2 Bonnie Tyler version
2.3 Julian Austin version
3 Personnel (original Meat Loaf version)
4 References
5 External links
Subsequent versions[edit]
Country music group Nitty Gritty Dirt Band performed a cover for their 1984 album Plain Dirt Fashion.
Canadian country music singer Julian Austin performed a cover for his 1997 debut album What My Heart Already Knows and reached No. 10 on the RPM Top Country Tracks charts.
Charts[edit]
Meat Loaf version[edit]
Weekly charts[edit]
Chart (1977-78)
Peak
position
Australia (Kent Music Report)[2] 11
Austria 11
Canada RPM Top Singles[3] 5
Canada RPM Adult Contemporary[4] 6
Irish Singles Chart 25
New Zealand [5] 9
UK Singles Chart 32
US Billboard Hot 100 11
US Billboard Adult Contemporary[6] 31
US Cash Box Top 100[7] 9
Year-end charts[edit]
Chart (1978)
Rank
Canada RPM Top Singles[8] 16
US Billboard Hot 100[9] 30
US Cash Box [10] 46
Bonnie Tyler version[edit]
Chart (1996)
Peak
position
US Hot Dance Maxi-Singles Sales (Billboard)[11]
4
Julian Austin version[edit]
Chart (1998)
Peak
position
Canada Country Tracks (RPM)[12] 20
Personnel (original Meat Loaf version)[edit]
Meat Loaf - lead vocals
Todd Rundgren - guitar, backing vocals
Kasim Sulton - bass guitar
Roy Bittan - piano
Roger Powell - synthesizer
John "Willie" Wilcox - drums
Rory Dodd - additional backing vocals
Strings arranged and conducted by Kenneth Ascher
References[edit]
Jump up ^ Meat Loaf - Two Out Of Three Ain't Bad (Japanese release), retrieved 2017-04-14
Jump up ^ Kent, David (1993). Australian Chart Book 1970–1992 (doc)|format= requires |url= (help). Australian Chart Book, St Ives, N.S.W. ISBN 0-646-11917-6.
Jump up ^ http://www.bac-lac.gc.ca/eng/discover/films-videos-sound-recordings/rpm/Pages/image.aspx?Image=nlc008388.4614a&URLjpg=http%3a%2f%2fwww.collectionscanada.gc.ca%2fobj%2f028020%2ff4%2fnlc008388.4614a.gif&Ecopy=nlc008388.4614a RPM Top Singles, July 8, 1978
Jump up ^ http://www.bac-lac.gc.ca/eng/discover/films-videos-sound-recordings/rpm/Pages/image.aspx?Image=nlc008388.6958&URLjpg=http%3a%2f%2fwww.collectionscanada.gc.ca%2fobj%2f028020%2ff4%2fnlc008388.6958.gif&Ecopy=nlc008388.6958 RPM Adult Contemporary, August 12, 1978
Jump up ^ https://charts.nz/showitem.asp?interpret=Meat+Loaf&titel=Two+Out+Of+Three+Ain%27t+Bad&cat=s
Jump up ^ Whitburn, Joel (1993). Top Adult Contemporary: 1961–1993. Record Research. p. 160.
Jump up ^ "Cash Box Top 100 7/01/78". tropicalglen.com.
Jump up ^ "Item Display - RPM - Library and Archives Canada". Collectionscanada.gc.ca. Retrieved 2017-04-05.
Jump up ^ "Top 100 Hits of 1978/Top 100 Songs of 1978". Musicoutfitters.com. Retrieved 2016-10-13.
Jump up ^ "Cash Box YE Pop Singles - 1981". Tropicalglen.com. 1981-12-26. Retrieved 2016-10-19.
Jump up ^ Billboard Vol. 108, No. 22. Nielsen Business Media, Inc. 1996. p. 29. ISSN 0006-2510.
Jump up ^ "Top RPM Country Tracks: Issue 3492." RPM. Library and Archives Canada. March 23, 1998. Retrieved October 14, 2013.
External links[edit]
Lyrics of this song at MetroLyrics
show
v
t
e
Bat Out of Hell
Meat Loaf
Songs by Jim Steinman
Produced by Todd Rundgren
Side 1
"Bat Out of Hell"
"You Took the Words Right Out of My Mouth"
"Heaven Can Wait"
"All Revved Up with No Place to Go"
Side 2
"Two Out of Three Ain't Bad"
"Paradise by the Dashboard Light" (duet with Ellen Foley)
"For Crying Out Loud"
Related articles
Bat Out of Hell II: Back into Hell
Bat Out of Hell III: The Monster Is Loose
Bat Out of Hell: Live with the Melbourne Symphony Orchestra
Cleveland International Records
show
v
t
e
Meat Loaf
Studio albums
Stoney & Meatloaf (1971)
Bat Out of Hell (1977)
Dead Ringer (1981)
Midnight at the Lost and Found (1983)
Bad Attitude (1984)
Blind Before I Stop (1986)
Bat Out of Hell II: Back into Hell (1993)
Welcome to the Neighbourhood (1995)
Couldn't Have Said It Better (2003)
Bat Out of Hell III: The Monster Is Loose (2006)
Hang Cool Teddy Bear (2010)
Hell in a Handbasket (2011)
Braver Than We Are (2016)
Live albums
Live at Wembley (1987)
Live Around the World (1996)
VH1: Storytellers (1999)
Bat Out of Hell: Live with the Melbourne Symphony Orchestra (2004)
3 Bats Live (2008)
Compilations
Hits Out of Hell (1984)
Heaven & Hell (1989)
Heaven Can Wait – The Best Ballads of Meat Loaf Vol. 1 (1996)
The Very Best of Meat Loaf (1998)
Videography
Live
Hits Out of Hell
Bad Attitude – Live!
Bat Out of Hell II: Picture Show
VH1: Storytellers
To Hell and Back
Bat Out of Hell: Live with the Melbourne Symphony Orchestra
3 Bats Live
Meat Loaf: In Search of Paradise
Tours
The Seize the Night Tour (2006–07)
The Casa de Carne Tour (2008)
The Hang Cool Tour (2010–11)
Guilty Pleasure Tour (2011)
Mad, Mad World Tour (2012)
Last at Bat Tour (2013)
Singles
"What You See Is What You Get"
"It Takes All Kinds of People"
"You Took the Words Right Out of My Mouth"
"Paradise by the Dashboard Light"
"Two Out of Three Ain't Bad"
"Bat Out of Hell"
"I'm Gonna Love Her for Both of Us"
"Dead Ringer for Love"
"Read 'Em and Weep"
"Peel Out"
"In Europe '82"
"If You Really Want To"
"Razor's Edge"
"Midnight at the Lost and Found"
"Modern Girl"
"Nowhere Fast"
"Rock 'n' Roll Mercenaries"
"Blind Before I Stop"
"Special Girl"
"I'd Do Anything for Love (But I Won't Do That)"
"Life Is a Lemon and I Want My Money Back"
"Rock and Roll Dreams Come Through"
"Objects in the Rear View Mirror May Appear Closer Than They Are"
"I'd Lie for You (And That's the Truth)"
"Not a Dry Eye in the House"
"Runnin' for the Red Light (I Gotta Life)"
"A Kiss Is a Terrible Thing to Waste"
"No Matter What"
"Is Nothing Sacred"
"Did I Say That?"
"Couldn't Have Said It Better"
"Man of Steel"
"It's All Coming Back to Me Now"
"Cry Over Me"
"Los Angeloser"
Related articles
Discography
Neverland Express
Stoney & Meatloaf
Ted Nugent
Jim Steinman
Patti Russo
Shaun Murphy
Amanda Aday
Pearl Aday
show
v
t
e
Bonnie Tyler songs
Discography
The World Starts Tonight
"Lost in France"
"More Than a Lover"
Natural Force
"It's a Heartache"
"Living for the City"
"Heaven"
"(You Make Me Feel Like) A Natural Woman"
"Here Am I"
Diamond Cut
"Too Good to Last"
"My Guns Are Loaded"
Goodbye to the Island
"I'm Just a Woman"
"Sometimes When We Touch"
"A Whiter Shade of Pale"
"Sitting on the Edge of the Ocean"
"I Believe in Your Sweet Love"
Faster Than the Speed of Night
"Have You Ever Seen the Rain?"
"Faster Than the Speed of Night"
"Total Eclipse of the Heart"
"Straight from the Heart"
Secret Dreams and Forbidden Fire
"If You Were a Woman (And I Was a Man)"
"Loving You's a Dirty Job but Somebody's Gotta Do It"
"Band of Gold"
"Lovers Again"
"Holding Out for a Hero"
Hide Your Heart
"Hide Your Heart"
"Don't Turn Around"
"Save Up All Your Tears"
"To Love Somebody"
"The Best"
Bitterblue
"Against the Wind"
"Bitterblue"
"Where Were You"
Angel Heart
"Fools Lullaby"
"I Cry Myself to Sleep at Night"
Silhouette in Red
"You Are So Beautiful"
Free Spirit
"You're the One"
"Making Love Out of Nothing at All"
"Bridge over Troubled Water"
"Two Out of Three Ain't Bad"
All in One Voice
"Angel of the Morning"
"I Put a Spell on You"
"The Rose"
"He's the King"
"Silent Night"
Heart Strings
"Human Touch"
"Everybody Hurts"
"Amazed"
"Against All Odds (Take a Look at Me Now)"
"And I Am Telling You I'm Not Going"
"Lean on Me"
"In My Life"
"Learning to Fly"
"Right Here Waiting"
"I Still Haven't Found What I'm Looking For"
"I Can't Make You Love Me"
"Need Your Love So Bad"
"It's Over"
Simply Believe
"Si demain... (Turn Around)"
"Si tout s'arrête (It's a Heartache)
"Darlin'"
Wings
"Louise"
Rocks and Honey
"This Is Gonna Hurt"
"Believe in Me"
"What You Need from Me"
"Love Is the Knife"
"Mom"
Collaborations
"A Rockin' Good Way (to Mess Around and Fall in Love)"
"Islands"
"Vergiß Es (Forget It)"
"Amour Éternel (Eternal Flame)"
"Love's Holding On"
Other songs
"Sayonara Tokyo"
"Here She Comes"
"The Desert Is in Your Heart"
"I Don't Know How to Love Him"
show
v
t
e
Julian Austin
Studio albums
What My Heart Already Knows
Back in Your Life
Bulletproof
The Red and White
One for One
Notable singles
"Little Ol' Kisses"
"Diamond"
"Two Out of Three Ain't Bad"
"What My Heart Already Knows"
"Hard Time Loving You"
"Take the Money and Run"
"Forever Loving You"
Retrieved from "https://en.wikipedia.org/w/index.php?title=Two_Out_of_Three_Ain%27t_Bad&oldid=851475100"
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1977 singles
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text:
Canada men's national soccer team records and statistics
From Wikipedia, the free encyclopedia
Jump to: navigation, search
Canada's men's national soccer team and its players[1][2] hold the following competitive records.
All official records are updated and published annually in the Canada Soccer Yearbook (2008,[3] 2009,[4] 2010,[5] 2011[6]).
Contents
[hide]
1 Competitive record
1.1 Pan American Games record
1.2 NAFC Championship record
1.3 Olympic Games record
2 Team and Individual records
2.1 Team
2.2 Individual
3 Player records
3.1 Player awards
3.1.1 Note
4 All-time record against other nations
5 See also
6 References
Competitive record[edit]
World Cup record
Year
Round
GP
W
D*
L
GS
GA
1930 Did Not Enter - - - - - -
1934 Did Not Enter - - - - - -
1938 Did Not Enter - - - - - -
1950 Did Not Enter - - - - - -
1954 Did Not Enter - - - - - -
1958 Did Not Qualify - - - - - -
1962 Withdrew - - - - - -
1966 Did Not Enter - - - - - -
1970 Did Not Qualify - - - - - -
1974 Did Not Qualify - - - - - -
1978 Did Not Qualify - - - - - -
1982 Did Not Qualify - - - - - -
1986 Round 1 3 0 0 3 0 5
1990 Did Not Qualify - - - - - -
1994 Did Not Qualify - - - - - -
1998 Did Not Qualify - - - - - -
2002 Did Not Qualify - - - - - -
2006 Did Not Qualify - - - - - -
2010 Did Not Qualify - - - - - -
2014 Did Not Qualify - - - - - -
2018 Did Not Qualify - - - - - -
Total 1/21 3 0 0 3 0 5
CONCACAF Championship/Gold Cup
Total: 2 Titles
Year
Round
GP
W
D*
L
GS
GA
1963 Did Not Qualify - - - - - -
1965 Did Not Qualify - - - - - -
1967 Did Not Qualify - - - - - -
1969 Did Not Qualify - - - - - -
1971 Did Not Qualify - - - - - -
1973 Did Not Qualify - - - - - -
1977 Fourth place 5 2 1 2 7 8
1981 Fourth place 5 1 3 1 6 6
1985 Champions 4 2 2 0 4 2
1989 Did Not Qualify - - - - - -
1991 Round 1 3 1 0 2 6 9
1993 Round 1 3 0 2 1 3 11
1996 Round 1 2 1 0 1 4 5
1998 Withdrew - - - - - -
2000 Champions 5 3 2 0 7 3
2002 Third place 5 2 2 1 5 4
2003 Round 1 2 1 0 1 1 2
2005 Round 1 3 1 0 2 2 4
2007 Semi-Finals 5 3 0 2 9 5
2009 Quarter-Finals 4 2 1 1 4 3
2011 Round 1 3 1 1 1 2 3
Total 2 Titles 49 20 14 15 60 65
Confederations Cup record
Year
Round
GP
W
D*
L
GS
GA
1992 to 1999 Did Not Qualify - - - - - -
2001 Round 1 3 0 1 2 0 5
2003 to 2017 Did Not Qualify - - - - - -
Total 1/10 3 0 1 2 0 5
*Denotes draws include knockout matches decided on penalty kicks.
**Silver background color indicates that the tournament was held on home soil. Gold background color indicates that the tournament was won.
Pan American Games record[edit]
1951 – Did not enter
1955 – Did not enter
1959 – Did not enter
1963 – Did not enter
1967 – Fourth place
1971 – Fifth place
1975 – Round 2
1979 – Did not enter
1983 – Did not enter
1987 – Round 999
1991 – Round 1
1995 – Did not enter
1999 – Fourth place
2003 – Did not enter
2007 – Did not enter
NAFC Championship record[edit]
1947 – Did not enter
1949 – Did not enter
1990 – Champions
1991 – Third place
Olympic Games record[edit]
Host nation(s) / Year
Result
GP
W
D*
L
GS
GA
1900 Did Not Enter - - - - - -
1904 Champions 2 2 0 0 11 0
1908 – 1964 Did Not Enter - - - - - -
1968 – 1972 Did not Qualify - - - - - -
1976 Group Stage 2 0 0 2 2 5
1980 Did not qualify - - - - - -
1984 Quarter-finals 4 1 2 1 5 4
1988 – 2012 Did Not Qualify - - - - - -
2016 Did Not Qualify - - - - - -
Total
3/24
8
3
2
3
18
9
*Denotes draws including knockout matches decided on penalty kicks.
**Gold background color indicates that the tournament was won. Red border color indicates tournament was held on home soil.
Olympic medal record
Men's Football
1904 St Louis Team
Team and Individual records[edit]
As of October 6, 2016
Team[edit]
First game
2–3 loss to Australia June 7, 1924
First win
1–0 win against Australia June 14, 1924
Biggest win
7–0 vs Saint Lucia October 7, 2011
Biggest loss
0–8 vs Mexico June 18, 1993
Most goals scored in one game
7 – vs Saint Lucia on October 7, 2011
Longest winning streak
6 games – from February 20, 2000 vs Mexico to June 4, 2000 vs Cuba
Longest undefeated streak
15 games – from September 2, 1999 vs Jamaica to June 4, 2000 vs Cuba
Longest losing streak
8 games – from October 31, 1974 vs Poland to August 10, 1975 vs Hungary
Longest winless streak
16 games – from October 16, 2012 vs Honduras to May 27, 2014 vs Moldova
Longest goalless streak
959 minutes – from March 22, 2013 vs Japan to May 23, 2014 vs Bulgaria
Longest clean sheet streak
567 minutes – from January 19, 2015 vs Iceland to July 11, 2015 vs Jamaica
Individual[edit]
Most appearances
89 – Julian de Guzman
Longest international career
22 years – Pat Onstad from 1988 to 2010
Most goals
22 – Dwayne De Rosario
Most goals in one game
3 – John Catliff vs Jamaica in 1988, Alex Bunbury vs Bermuda in 1992 and Simeon Jackson vs Saint Lucia in 2011
Most goals in one year
7 – Igor Vrablic in 1985
Most consecutive games with at least one goal
4 – Dale Mitchell and Ali Gerba
Most consecutive games with at least two goals
2 – Tom Kouzmanis vs Jamaica and Trinidad and Tobago in 1995, and Ali Gerba vs Saint Vincent and the Grenadines (twice) in 2008
Most braces (2 goals in one game)
5 – Dale Mitchell vs New Zealand in 1980, Trinidad and Tobago in 1981, Guatemala in 1985 and 1988, and Honduras in 1991
Most penalty kicks scored
7 – Dwayne De Rosario
Least amount of time in between two consecutive goals
1 minute – Tom Kouzmanis vs Jamaica in 1995
Longest amount of time between first and last goals
12 years and 11 months – Dwayne De Rosario vs South Korea in 2002 and Iceland in 2015
Most appearances before scoring a first international goal
49 – Nik Ledgerwood vs El Salvador in 2016
Fastest goal from kickoff
4th minute – Adrian Serioux vs Honduras in 2008, Ali Gerba vs Guatemala in 2009, and Igor Vrablic vs Indonesia in 1986, Tesho Akindele vs Dominica in 2015
Latest goal in regulation time
93rd minute – Gavin McCallum vs Venezuela in 2010, Tosaint Ricketts vs Puerto Rico in 2011
Latest goal in extra time
120th minute – John Catliff vs Mexico in 1990
Most goals in a debut
2 – Art Hughes, Dale Mitchell, Nick Papadakis, Tom Kouzmanis
Player records[edit]
As of March 24, 2018.
Most caps
Rank
Name
Career
Caps
Goals
1
Julian de Guzman 2002–2016 89 4
2
Paul Stalteri 1997–2010 84 7
3
Randy Samuel 1983–1997 82 0
4
Dwayne De Rosario 1997–2015 81 22
5
Atiba Hutchinson 2003– 78 6
Mark Watson 1991–2004 78 3
7
Lyndon Hooper 1986–1997 67 3
8
Alex Bunbury 1986–1997 66 16
9
Nick Dasovic 1992–2004 63 2
Kevin McKenna 2000–2012 63 11
Bold notes player is still active.
Most goals
Rank
Name
Career
Caps
Goals
1
Dwayne De Rosario 1997–2015 81 22
2
Dale Mitchell 1980–1993 55 19
3
John Catliff 1984–1994 44 18
4
Tosaint Ricketts 2011– 58 16
Alex Bunbury 1986–1997 66 16
6
Ali Gerba 2005–2011 31 15
7
Igor Vrablic 1984–1986 36 12
8
Carlo Corazzin 1994–2004 59 11
Kevin McKenna 2000–2012 63 11
10
Tomasz Radzinski 1995–2009 46 10
Paul Peschisolido 1992–2004 53 10
Bold notes player is still active.
Player awards[edit]
Awarded by the Canadian Soccer Association.[7]*
See: Canadian Players of the Year
Note[edit]
*The Canadian Players of the Year are determined by voting from Canadian media and coaches, save for 2002 when it was determined by fan voting, and 2005 and 2006 when it was determined by fans and media voting. The Canadian U-20 Player of the Year was determined by fans and media voting until 2007, from when it was determined by the voting of Canadian coaches.
All-time record against other nations[edit]
As of March 24, 2018[8]
Team
GP
W
D
L
GF
GA
Algeria 1 0 0 1 0 1
Argentina 1 0 0 1 0 5
Armenia 1 0 0 1 1 3
Australia 9 3 1 5 11 15
Austria 1 1 0 0 2 0
Azerbaijan 1 0 1 0 1 1
Barbados 1 1 0 0 1 0
Belarus 2 1 0 1 1 2
Belgium 1 0 0 1 0 2
Belize 4 3 1 0 12 1
Bermuda 9 5 4 0 17 4
Brazil 4 0 2 2 4 8
Bulgaria 1 0 1 0 1 1
Cameroon 1 0 0 1 0 2
Chile 4 1 1 2 2 4
China PR 3 1 0 2 6 8
Colombia 3 1 0 2 2 4
Costa Rica 20 4 8 8 14 20
Cuba 11 6 3 2 13 9
Curaçao 1 1 0 0 2 1
Cyprus 2 1 1 0 1 0
Czech Republic 2 0 0 2 1 7
Denmark 3 0 0 3 0 7
Dominica 2 2 0 0 6 0
East Germany 3 0 0 3 1 12
Ecuador 4 0 1 3 4 8
Egypt 2 0 0 2 0 4
El Salvador 17 8 4 5 20 17
England 1 0 0 1 0 1
Estonia 2 0 0 2 1 4
Faroe Islands 2 1 0 1 1 1
Finland 1 0 0 1 2 3
France 1 0 0 1 0 1
French Guiana 1 1 0 0 4 2
Germany 2 0 0 2 1 6
Ghana 2 1 1 0 3 2
Greece 4 0 1 3 0 5
Guadeloupe 2 1 0 1 2 2
Guatemala 14 10 2 2 22 10
Haiti 10 7 2 1 15 8
Honduras 23 7 5 11 28 40
Hong Kong 1 1 0 0 3 1
Hungary 4 0 0 4 3 10
Iceland 3 0 2 1 3 4
Indonesia 1 1 0 0 4 0
Iran 3 1 0 2 1 2
Iraq 1 0 0 1 1 6
Italy 1 0 0 1 0 2
Jamaica 21 9 6 6 24 17
Japan 2 0 0 2 1 5
Libya 1 1 0 0 4 2
Luxembourg 2 2 0 0 3 0
Macedonia 2 1 0 1 1 3
Malaysia 1 1 0 0 5 0
Malta 3 0 1 2 1 4
Martinique 4 1 2 1 4 4
Mauritania 3 1 1 1 4 1
Mexico 31 4 8 19 19 68
Morocco 3 0 1 2 3 8
Moldova 1 0 1 0 1 1
Netherlands 1 0 0 1 0 3
New Zealand 7 5 1 1 16 5
Nigeria 1 1 0 0 3 1
Northern Ireland 3 2 1 0 4 1
North Korea 2 0 1 1 0 2
Panama 10 3 6 1 8 6
Paraguay 1 0 1 0 0 0
Peru 2 1 0 1 3 3
Poland 6 0 0 6 4 20
Portugal 2 0 1 1 2 5
Puerto Rico 3 2 1 0 6 0
Republic of Ireland 1 0 0 1 0 3
Saint Kitts and Nevis 2 1 1 0 4 0
Saint Lucia 2 2 0 0 11 1
Saint Vincent and the Grenadines 2 2 0 0 7 1
Saudi Arabia 1 0 0 1 0 2
Scotland 6 0 1 5 3 14
Singapore 2 2 0 0 2 0
Slovenia 1 0 0 1 0 1
South Africa 1 0 0 1 0 2
South Korea 5 2 1 2 4 5
Spain 2 0 0 2 1 4
Soviet Union 1 0 0 1 0 2
Suriname 1 1 0 0 2 1
Switzerland 1 1 0 0 3 1
Trinidad and Tobago 11 7 2 2 16 12
Tunisia 1 0 0 1 0 2
Turkey 2 0 0 2 1 6
Ukraine 1 0 1 0 2 2
Uruguay 1 0 0 1 1 3
United States 34 8 12 14 33 52
Uzbekistan 1 1 0 0 2 1
Venezuela 2 0 2 0 3 3
Wales 3 1 0 2 2 4
See also[edit]
Canada men's national soccer team
List of Canada international soccer players
References[edit]
Jump up ^ CanadaSoccer.com | Official Site of the Canadian Soccer Association
Jump up ^ "Canadian Scoring Stats..."
Jump up ^ 2008 Yearbook
Jump up ^ 2009 Yearbook
Jump up ^ 2010 Yearbook
Jump up ^ 2011 Yearbook
Jump up ^ "2008 Yearbook"
Jump up ^ "Canada national football team statistics and records: All-time record". 11v11.com.
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