Mental maths – A guide to effective mental calculation

Warning: I may use English styles of notation. This includes using commas as a way to divide up the thousands in long numbers (i.e. 32,000 = thirty-two thousand), I will use full stops (periods) as decimal points.

Calculating things in your head can be a difficult task. If you can’t remember what you’ve worked out or simply don’t know how to solve a problem then it can be very challenging and frustrating. I’m going to try and give a few tips on how to do it more easily. My own mental calculation skills are below my general maths ability due to problems with short term memory, but with a few shortcuts I can often calculate things scarily fast.

[edit]Addition

A useful trick when adding lots of small numbers is to clump together the ones that add up to multiples of 10. For example, if you have to add 2 + 3 + 5 + 7 + 9 + 11 + 8, that can be rearranged as (3 + 7) + (9 + 11) + (2 + 8) + 5 = 10 + 20 + 10 + 5 = 45.

This method is also useful when performing column addition with more than two numbers. For example, in the problem:

56

35

47

21

12

32

+23

—

Column addition is generally performed by adding the digits in the ones place, carrying them over, and then adding the digits in the tens place., and so on. A way to make this task easier is to group the digits in the ones place in groups of ten, and mark them on your paper like this:

5 6

3 5

4 7\

2 1 \

1 2 — 10

3 2 /

+2 3/

—

Similarly the 6, 2, and 2 would be crossed off, yielding another 10. Therefore the digits in the ones place add up to 10+10+5+1 (what’s left) or 26.

[edit]Subtraction

A useful trick when subtracting numbers is to begin with the smaller value and mentally skip your way up the difference, with jumping points at recognizable boundaries, such as powers of 10. For example, to subtract 67 from 213 I would start with 67, then add 3 + 30 + 100 + 13. Try this once and you see how easy it is. Sounding out your thoughts it would be “three, thirty-three, one hundred thirty-three plus the remaining 13 is one hundred forty-six”.

[edit]Subtraction from Numbers consisting of 1 followed by zeros: 100; 1,000; 10,000; etc

For example 1,000 – 258 We simply subtract each digit in 258 from 9 and the last digit from 10.

2 5 8

from 9 from 9 from 10

7 4 2

So the answer is 1,000 – 258 = 742

And thats all there is to it!

This always works for subtractions from numbers consisting of a 1 followed by zeroes: 100; 1000; 10,000 etc.

A second method is to break up the number that you are subtracting. So instead of doing 1000-258 you would do 1000-250 and then subtract 8.

Another way of easily thinking of this method is to always subtract from 999 if subtracting from 1,000, and then adding 1 back. Same for 10,000, subtract from 9999 and add 1. For example, 1000-666 = 999 – 666 + 1= 333 + 1 = 334

Similarly 10,000 – 1068 = (9999-1068)+1 = (8931)+1 =8932 So the answer is 10,000 – 1068 = 8932

For 1,000 – 86, in which we have more zeros than figures in the numbers being subtracted, we simply suppose 86 is 086. So 1,000 – 86 becomes 1,000 – 086 = 914

[edit]Multiplication

When multiplying it is very important to pick the correct sums to do. If you multiply 251 by 323 straight off it can be very difficult, but it is actually a very easy sum if approached in the right way. 251×3 + 251×20 + 251×300 is a scary prospect, so you have to work out the simplest method.

[edit]Rounding

One of the first things to do is to look if the numbers are near anything easy to work out. In this example there is, very conveniently, the number 251, which is next to 250. So all you have to do is 323×250 + 323 – much easier, but 323×250 still doesn’t look too simple. There is, however, an easy way of multiplying by 250 which can also apply to other numbers. You multiply by 1000 then divide by 4. So 323×1000 = 323,000, divide by two and you get 161,500, divide by 2 again and you get 80,750. Now this may not seem easy, but once you’ve gotten used to it, dividing by four (or other low numbers) in that way becomes natural and takes only a fraction of a second. 80,750+323 = 81,073 , so you’ve got the answer with a minimum of effort compared to what you would otherwise have done. You can’t always do it this easily, but it is always useful to look for the more obvious shortcuts in this style.

An even more effective way in some circumstances is to know a simple rule for a set of circumstances. There are a large number of rules which can be found, some of which are explained below.

[edit]Factoring

If you recognize that one or both numbers are easily divisible, this is one way to make the problem much easier. For example, 72 x 39 may seem daunting, but if taken as 8 x 9 x 3 x 13, it becomes much easier.

First, rearrange the numbers in the hardest to multiply order. In this case, I’d go with 13 x 8 x 9 x 3. Then multiply them one at a time.

13 x 8 = 10 x 8 + 3 x 8 = 80 + 24 = 104

104 x 9 = 936

936 x 3 = 2808 that would equal another number

Which would equal up to a whole new Faction

[edit]Multiplication by 11

To multiply any 2-digit number by 11 we just put the total of the two digits between the 2 figures.

for example: 27×11 can be written as [2][2+7][7] Thus, 27×11=297

Another example: 33×11 can be written as [3][3+3][3] Thus, 33×11=363. To visualise:

330

+ 33

—-

363

Carry=overs:

77 x 11 = 847 This involves a carry figure because 7 + 7 = 14 we get 77 x 11 = [7][14][7]. We add the 1 from 14 as carry over to 7 and get 77×11=847

Similarly, 84×11 can be written as [8][8+4][4]=[8][12][4]. The 1 from 12 carries over, giving 84×11=924

For 3 digit numbers multiplied by 11:

254 x 11 = 2794 We put the 2 and the 4 at the ends. We add the first pair 2 + 5 = 7. and we add the last pair: 5 + 4 = 9. So we can write 254 x 11 as [2][2+5][5+4][4] i.e. 254×11=2794

Similarly, 909×11 can be written as [9][9+0][0+9][9] i.e. 909×11=9999

[edit]Same First Digit, Second Digits Add to 10

Let’s say you are multiplying two numbers, just two two-digit numbers for now (though the rules could be adapted for others) which start with the same digit and the sum of their unit digits is 10. For example, 87×83 (sum of unit digits: 7+3=10). You multiply the first digit by one more than itself (8×9 = 72). Then multiply the second digits together (7×3 = 21). Then stick the first answer at the start of the second to get the answer (7221). A simple proof of how this works is given in the Wikipedia article on Swami Bharati Krishna Tirtha’s Vedic mathematics.

If the result from the multiplication of the unit digits is less than 10, simply add a zero in front of the number (i.e., 9 becomes 09). For example, 59×51 is equal to [5×6][9×1] which equals [30][09]. Thus 59×51 = 3009.

[edit]Squaring a Number That Ends with 5

This is a special case of the previous method. Discard the 5, and multiply the remaining number by itself plus one. Then tack on a 25 (which as in the previous section, is 5×5). For example, 65×65. Discarding the 5 from 65 leaves us with 65 = 6. Multiplying 6 by itself plus one gives us 42 (6×7 = 42). Tacking on a 25 yields 4225, so 65×65=4225. For example, 45×45 can be written as [4×5][5×5]thus 45×45 = 2025

[edit]Squaring a two-digit number

Rather than doing 142 or 472 as 14×14 or 47×47, the alternative is:

142

= 10 x 1(14 + 4) + (4 x 4)

= 10(18) + 16

= 180 + 16

= 196

In other words, add what’s in the ones place to the number, multiply it by what was originally in the tens place (sometimes you’ll get a sum with the next number up in the tens i.e. 47 + 7 = 54 so use 4 not 5 in this example) tack a zero at the end, then add the square of the ones place. So:

472

= 10 × 4(47 + 7) + (7 × 7)

= 10 × 4(54) + 49

= 10 × 216 + 49

= 2160 + 49 = 2209

So now we know that 472 is 2209.

When squaring two digit numbers that are only 1 away from a number ending in zero you can also use the basic algebraic formulas (A^2)-(A-1)^2 = 2A-1 and (A+1)^2 – (A^2) = 2A + 1. For example when squaring 99 you can set 100 as A then: that’s not true because they explained it the wrong way 100^2 = 10000 2 * 100 = 200 So the answer is (10000 – 200) + 1 = 9801

To square 91 use the second formula. Then: 90^2 = 8100 2*90 = 180 So the answer is 8100 + 180 + 1 = 8281

[edit]Squaring a number when you know the square of a number adjacent or in proximity

This is useful if you want to quickly calculate the square of a number when you know the square of the adjacent number. For example, take the square of 46, using the “5” rule above you know that 45 squared is 2025. Leverage this number and add 45+46 (91) to 2025, which equals 2116. While adding 91 to 2025 in your head isn’t exactly easy, it is certainly easier than trying to calculate the square of 46 directly. Doing this with an adjacent known square that is below is a bit more challenging depending on how you feel about doing subtraction in your head. For subtraction, using 45 as our base, and trying to figure our what 44 squared is, we would take the known value of 2025 subtract 44 and 45 to get 1936. This can be leveraged to try to determine squares that are not directly adjacent to the known square, but it gets a bit more complex (in your head!). Symbolically: if b=a+1 and a and b are integers then b2=a2+|a|+|b|.

[edit]Just Over 100

This trick works for two numbers that are just over 100, as long as the last two digits of both numbers multiplied together is less than 100. For example, for 103 x 124, 3 x 24 = 72 100, so it will not.

If the first test works, then the answer is:

1[sum of last two digits][product of last two digits]

Examples:

108 x 109 = 1[8+9][8×9] = 1[17][72] = 11,772

105 x 115 = 1[5+15][5×15] = 1[20][75] = 12,075

132 x 103 = 1[32+3][32×3] = 1[35][96] = 13,596

If the addition or multiplication of the last 2 digits < 10, then add a 0 infront of the number, example if the addition is 4, it should be 04. Example shown below:

102 x 103 = 1[2+3][2×3]=1[05][06]=10,506

This trick works for numbers just over 200, 300, 400, etc. with one simple change:

[product of first digits][(sum of last two digits) x first digit][product of last two digits]

Examples:

215 x 204 = [2×2][(15+4)x2][15×4] = [4][19×2][60] = [4][38][60] = 43,860

If the addition or multiplication of the last 2 digits < 10, then add a 0 infront of the number, example if the addition is 4, it should be 04. Example shown below:

201 x 202 = [2×2][(2+1)x2][2×1]=[4][06][02]= 4,0602

For numbers just over 1000, 2000, etc., use the following:

[product of first digits]0[(sum of last two digits) x first digit]0[product of last two digits]

Examples:

2008 x 2009 = [2×2]0[(8+9)x2]0[8×9] = [4]0[17×2]0[72] = [4]0[34]0[72] = 4,034072

2008 x 2009 = 4,034,072

For each order of magnitude (x10), add two zeroes to the middle.

[edit]Division

Again there are many possible techniques, but you can make do with the following or research your own. All numbers are the products of primes (you can make them by multiplying together prime numbers). If you are dividing you can divide by all the prime products of the number you are dividing by to get the answer. This means that 100/24 = (((100/2)/2)/2)/3. Although this means you have a lot more stages to do they are all much simpler. 100/2 = 50 , 50/2 = 25 , 25/2 = 12.5 , 12.5/3 = 45/30 = 41/6 = 4.166666666recurring

Also, another helpful trick is, when you have to muliply and then divide by a number, always divide first, until you've reached numbers that are relatively prime, and then multiply. This keeps numbers from being too large. For example, if you must do (18 * 115)/15, it is much easier to divide 115 by 5 and 18 by 3, and then multiply them together to get 23 * 6 = 138.

[edit]Multiply by the Reciprocal

Division is equivalent to multiplying by the reciprocal. For instance, division by 5 is the same as multiplication by 0.2 (1/5=0.2). To multiply by 0.2, simply double the number and then divide by 10.

[edit]Division by 7

The number 1/7 is a special number, equal to . Note that there are six digits that repeat, 142857. A beautiful thing happens when we consider integer multiples of this number:

Note that these six fractions of seven contain the same six digits repeating in the same order ad infinitum, but starting with a different number. But how is this useful when dividing by seven? Consider the problem 207/7. First, we can convert this to 200/7 + 7/7. We know that 7/7 equals one, so the answer will be 200/7 + 1. But what is 200/7? It is simply 2/7 times 100, and from the above, we know that , so by moving the decimal point, we know that . All that remains is to add the one from 7/7, giving us .

[edit]Division by 9

The fraction 1/9 and its integer multiples are fairly straightforward – they are simply equal to a decimal point followed by the one-digit the numerator repeating to infinity:

To solve a problem such as 367/9, we reduce it to

First add . Then add .

[edit]Division by 11

The fraction 1/11 and its integer multiples are fairly straightforward – they are simply equal to a decimal point followed by the product of nine and the the integer multiple repeating to infinity:

[edit]Estimation

The best way to make estimation quickly in mental math is to round to one or two significant digits (that is, round it to the nearest place of the highest order(s) of magnitude), and then proceed with typical operations. Thus, 1242 * 15645 is approximately equal to 1200 * 16000 = 19200000, which is reasonably close to the correct answer of 19431090. In certain cases, one can even round to simply the nearest power of ten (which is useful when making estimations with much error and large numbers).

[edit]Range Search

It is sometimes easier to make a calculation in the opposite direction from the one you want, and this can be used to quickly estimate the value you want.

Square root is a good example. It is easier to square numbers than to take the square root. So you take any number that will square to a little larger than the one you want, take another that squares to less than the one you want and use an average of the two.

The trick now is to apply a general technique ( the Bisection Method ).

We create a new estimate using the average of our first two numbers. Square this. If it's value is higher than we want, we use it as the upper value of our range. If it is lower than we want, we use it as the lower value of our range.

We now have a new range that must contain the square root we want. We can apply the same process again to get a more accurate value ( this is known as Iteration ).

This technique is widely applied in computing, but also very handy for some mental mathematics.

[edit]Other mental maths

Perhaps one of the more useful tricks to mental math is memorization. Although it may seem an annoyance to need to memorize certain math facts, such as perfect squares and cubes (especially powers of two), prime factorizations of certain numbers, or the decimal equivalents of common fractions (such as 1/7 = .1428…). Many are simple, such as 1/3 = .3333… and 2^5 = 32, but speed up your calculations enormously when you don't have to do the division or multiplication in your head. For example, trying to figure out 1024/32 is much easier knowing that that is the same as 2^10/2^5, or which, subtracting exponents, gives 2^5, or 32. Many of these are memorized simply by frequent use; so, the best way to get good is much practice.

It's a good idea to memorize 3 x 17 = 51. We can extend this to 6 x 17 = 102. If we round these numbers 3 x 17 is approx 50, 6 x 17 is approx 100, 9 x 17 is approx 150, and so on. These are very helpful in estimating since 3, 6, 9, 50, 100 … are common numbers.