Mental Math

Mental Math - A Guide to Effective Mental Calculations
''Note about Notation: This book generally uses the English/U.S. styles of notation. This includes using commas as a way to divide up the thousands in long numbers (e.g. 32,000 = thirty-two thousand), it 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. But by learning and practicing the methods of using mathematical patterns, you can dramatically improve the speed and accuracy of your arithmetic. These methods are often called "High Speed Mental Math". Mental Math is a valuable skill to have, even in the computerized world we live in:
 * 1) With good mental math skills you can save yourself time by not needing to pull out a calculator (or cell phone) every time you want to do a task.
 * 2) Mental math skills will improve your ability to estimate results, thus having a better ability to catch errors from computer-derived results.  For example, while a calculator will generally give the right answer, based upon what was typed in, if you accidentally typed the wrong number, you might not catch your error if you didn't have good mental arithmetic skills.

Addition
The foundation of all arithmetic is addition, also known as summing. Similar to all mental math calculations, you can improve your ability to add numbers by learning to use some basic patterns.

Changing the Order of Addition
Often when looking for a pattern to help you quickly do an addition problem, it can help to change the order that you add things. For example, 8+1 is the same as 1+8, and in both cases you just count up one from 8, to get the answer 9. If you get stumped on an addition problem, try changing the order of which one you add first, and see if it helps.

Adding Zero, One, or Two
Unless you are completely new to addition, you surely know the pattern for adding I zero. Anything plus zero is equal to the original number. Thus when you see a zero in an addition problem, you can basically ignore it, as it won't have any effect upon the answer. For those who are interested in math trivia, this property of zero not changing things when you add it, is called the "identity property" in arithmetic. Also, keep in mind, that you can only ignore zero in addition, and sometimes in subtraction. In multiplication and division, having a zero in a problem always changes things. You probably also know the rule of adding 1, which is simply count up one number. This can also be used quickly with 2, where you can either count up 2 numbers, or if you know your even and odd numbers well, you can skip to the next even or odd. To understand using the even and odd pattern, you may be familiar with the cheer that has the even numbers: "2,4,6,8, who do we appreciate?" If we add 6+2 we will go up to the next even number, which is 8. Similarly, the odd numbers go 1,3,5,7 so 2+7 will be the next odd number, which is 9.

Adding Nine or Eight: Counting Down from 10
Another arithmetic pattern you surely know is how to add 10 to a number. An example of this is 2+10 is 12, or 6+10 is 16. We can use this pattern to help us add 9 or 8. Since 9 is one less than 10, you can always add 9 to something by adding 10 to the number instead and then counting down one. For instance to find 9+7, you can add 10+7, which is 17, and then count down to 16. Similarly with adding 8 to a number, you can also add 10, and then count down 2 numbers. So for instance, 8+7 can be found by adding 10+7 which is 17, and then counting down 2, which is 15. If you are good with knowing your even and odd numbers, you can also use a similar pattern as was explained with adding 2, by just going down one even or odd number. or you could say when it's adding 9 you know it's going to a higher number for instance 9+8 you know it's going to be a teen so subtract one 9+8 = 1 now subtract one from 8 to get 7 now put them next to each other to get 17.

Doubling and Nearly Doubling
You may already be good at doubling numbers, such as 2+2. When doubling a number, you are doing the same thing as multiplying by 2. This also means, that if you have learned to count by numbers, such as counting by 4's, you know that it is 4, 8, 12, 16... And thus 4+4 is 8. Once you have become good at doubling, you can use this knowledge to add numbers that are nearly double of each other, by just counting up or down by one. So for instance, 8+7 can be found by adding 8+8, which is 16, and then counting down by 1, which is 15. (If it is easier, you could also have done this by doing 7+7, which is 14, and then counting up 1 to 15.)

Adding Fives
This technique is a little trickier than the others we have so far talked about, but with some practice it might help you out. When adding 5 to another number, the goal is to "find the five inside" the other number, and then add or subtract what is left. For instance when adding 5+8, you could say that 8 is 5+3, so 5+8 is 5+5+3, or 13. Don't worry if you can't get this method, because most of the time you can use one of the other methods that have been taught to find the answer.

To Ten or Close
It is useful to memorize the patterns that add to 10 which are: By knowing these patterns, if you see a pattern a little different you will know to go one higher or lower. For instance, 4+7 has the 7 one higher than 6, so this must add up to 11, because 4+6=10 so 4+6+1=11.
 * 0+10
 * 1+9
 * 2+8
 * 3+7
 * 4+6
 * 5+5

Summing Groups of Numbers by Finding Numbers that Add to Multiples of 10
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.

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". 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.

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

Note that if the number being subtracted (the minuend) ends in a series of zeros, you subtract the last nonzero digit from 10, and leave the trailing zeros alone. E.g.: 10,000 - 5920 = {9-5}{9-9}{10-2}0 = 4080. This always works for subtractions from numbers consisting of a 1 followed by zeroes: 100; 1000; 10,000 etc. 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, 1000-555 =   999 - 555 + 1= 444 + 1 = 445 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

Bigger Numbers using the 9s Trick (10's Complement Addition)
The above technique can be extended to arbitrarily long numbers in the following way: subtract the minuend from the next power of ten greater than the subtrahend using the nines trick above, then add the result to the subtrahend instead, but don't carry the final '1'.

Example: 1,254,953-34,064 34,064 becomes 9,965,936 via the nines trick (notice we pad the beginning with 9's to make it the same length as the subtrahend). Now add: 11  1 1  1,254,953 +9,965,936  1,220,889 And again notice that we do not include the final carried '1' on the left in our result.

Subtracting Left-to-Right
Normal subtraction is performed right-to-left, borrowing from the left when a borrow is needed. However, we read left to right, and numbers are read out loud to us in this same order, so it is convenient to be able to subtract left-to-right as well. We can do this by going ahead and pre-borrowing by default and then adding back a carry (if there is one) when we move on to the next column if it turns out we didn't need to borrow. This technique also cuts down on the amount of mental work one must do to propagate a borrow from right to left.

Example: 9543 -1992 First preborrow from the 9, leaving 8 and subtract the thousands place: 1 8543 -1992  7 Now preborrow from the 15 in the hundreds place, leaving a 14, and subtract: 11 8443 -1992  75 Now preborrow from the 14 in the tens place, leaving a 13, and subtract: 111 8133 -1992  754 Last column, we have nothing to borrow from, so we just subtract. In this case, we don't get a single digit, so we carry it back to the tens place: 111 8133 -1992  7551 And we're done! 7551 is the correct answer.

Patterns for 0, 1 and 10
Chances are you already know the pattern for multiplying by 0, 1 and 10. But in case you don't, anything times 0 is 0. Anything times 1, is itself still, and anything times 10 has a zero added to the end, so 29x10 is 290

Patterns for 2, 4, and 8
Multiplying by 2 is simply doubling a number, and so the pattern for doubling with addition is the same. Multiplying by 4 is doubling twice. So 12x4, can be found by going 12+12 which is 24 and then 24+24 which is 48. Similarly, multiplying by 8 is doubling 3 times, so 12x8 is 12+12 which is 24, then 24+24 which is 48, and then 48+48 which is 96. If you multiply by a quadratic (a number that can continuously divide in half until reaching 1, without involving fractions), then this method will always work.

Patterns for 9
Multiplying by 9 has a special pattern. When multiplying a single digit by 9, the answer will always start with a digit one less than the number, and then the other number will add to 9. This may sound complex, but lets look at an example. If we want to do 9x6, simply have your first digit being one less than 6, so we know the answer will start with a 5, then next digit must add up to 9, so the number that when added to 5 will equal 9, is 4, so the answer of 9x6 is 54. Another way to think of this method is you're multiplying the number by 10, then subtracting the number from the product. If you think of it this way, then this example becomes 10x6 = 60, then 60-6 = 54.

Patterns for 5
Any number multiplied by 5 will end in either a 5 or a 0. One way of finding the answer is to multiply by 10 first, and then divide your answer in half. Another is to count by fives.

"Number Neighbors"
If you don't know the answer to a problem, you may know the answer to a problem where one of the numbers is one more or less than the one in the problem. For instance, if you didn't know the answer to 7 x 6, you  might know 6 x 6 is 36, and then you could just add one more 6 to get to 42.

Multiplying Larger Numbers
When multiplying larger numbers 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. 251x3 + 251x20 + 251x300 is a scary prospect, so you have to work out the simplest method.

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 323x250 + 323 - much easier, but 323x250 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 323x1000 = 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.

For this multiplication, some would mentally find 251 × 323 = 250 × (320 + 3) + 1 × 323 (easy), since multiplying 250 × 320 is not hard: 25 × 32 = 25 × 4 × 8 = 100 × 8 = 800 (some would get this immediately), so 250 × 320 is 80,000. Thus, with everything in mind, 251 × 323 = 250 × 320 + 250 × 3 + 1 × 323 = 80,000 + 750 +323 = 81,073.

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. Which would equal up to a whole new Faction
 * 1) 13 x 8 = 10 x 8 + 3 x 8 = 80 + 24 = 104
 * 2) 104 x 9 = 936
 * 3) 936 x 3 = 2808 that would equal another number

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: 27x11 can be written as [2][2+7][7] Thus, 27x11=297 Another example: 33x11 can be written as [3][3+3][3] Thus, 33x11=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 77x11=847 Similarly, 84x11 can be written as [8][8+4][4]=[8][12][4]. The 1 from 12 carries over, giving 84x11=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. 254x11=2794 Similarly, 909x11 can be written as [9][9+0][0+9][9] i.e. 909x11=9999

Multiplication by 99, 999, 9999, etc
To multiply a number A by 99, you can multiply A by 100 and then subtract A from the result. When the number A is a two digit or one digit number, the result would be (A - 1) following by (100 - A). For example, when we multiply 65 by 99, we get 6435. Similarly, to multiply a number A by 999, you can multiply A by 1000 and then subtract A from the result. When the number A is a three digit, two digit or one digit number, the result would be (A - 1) following by (1000 - A). For example, when we multiply 611 by 999, we get 610389. This same idea can be used for multiplication by any large number consisting only 9s.

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.

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 5x5). For example, 65x65. Discarding the 5 from 65 leaves us with 6 5 = 6. Multiplying 6 by itself plus one gives us 42 (6x7 = 42). Tacking on a 25 yields 4225, so 65x65=4225. For example, 45x45 can be written as [4x5][5x5]thus 45x45 = 2025

Squaring a two-digit number
Rather than doing 142 or 472 as 14x14 or 47x47, 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 &times; 4(47 + 7) + (7 &times; 7) = 10 &times; 4(54) + 49 = 10 &times; 216 + 49 = 2160 + 49 = 2209 So now we know that 472 is 2209. When squaring two digit numbers that are only 1 higher from a number ending in zero, you can also use the basic algebraic formulas: (A+1)^2 - (A*2); or (A+1)(A) - A; or (A+1)^2 - 2(A+1) + 1. For example, when squaring 99: A + 1 = 100 100^2 = 10000 2 * 100 = 200 10000 - 200 = 9800 9800 + 1 = 9801 When squaring two digit numbers that are only 1 higher from a number ending in zero, you can also use the basic algebraic formulas: (A-1)^2 + (A*2); or (A-1)(A) + A; or (A-1)^2 + 2(A-1) + 1. For example, when squaring 91: 91 - 1 = 90 90^2 = 8100 2*90 = 180 8100 + 180 = 8280 8280 + 1 = 8281

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|.

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 this trick will work. For 117 x 112, 17 x 12 = 204 > 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: 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: This trick works for numbers just over 200, 300, 400, etc. with one simple change:
 * 108 x 109 = 1[8+9][8x9] = 1[17][72] = 11,772
 * 105 x 115 = 1[5+15][5x15] = 1[20][75] = 12,075
 * 132 x 103 = 1[32+3][32x3] = 1[35][96] = 13,596
 * 102 x 103 = 1[2+3][2x3]=1[05][06]=10,506
 * [product of first digits][(sum of last two digits) x first digit][product of last two digits]

Examples: 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: For numbers just over 1000, 2000, etc., use the following:
 * 215 x 204 = [2x2][(15+4)x2][15x4] = [4][19x2][60] = [4][38][60] = 43,860
 * 201 x 202 = [2x2][(2+1)x2][2x1]=[4][06][02]= 4,0602
 * [product of first digits]0[(sum of last two digits) x first digit]0[product of last two digits]

Examples: For each order of magnitude (x10), add two zeroes to the middle.
 * 2008 x 2009 = [2x2]0[(8+9)x2]0[8x9] = [4]0[17x2]0[72] = [4]0[34]0[72] = 4,034072
 * 2008 x 2009 = 4,034,072

Just beneath 100
This trick words for every number beneath 100, but is faster in the 90s. For example, 96*98:

The answer will be a four-digit number,

[xx][xx].

For the last two digits, it is the product of the differences between 100 and the number for each multiplier. Take the first number 96. 100-96=4. Then take the second number 98. 100-98=2. 4x2=8. 8 is going to be our last two digits so 08.

[xx][08]

The first two digits is any number minus the difference between 100 and the other number. Let's take the first number 96 for example. The other multiplier is 98. 100-98=2. So take 2 away from 96. 96-2=94. You will get the same two digits if you pick any of the two numbers.

[94][08]. So the answer to the equation is 9408.

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 multiply 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.

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.

Division by 7
The number 1/7 is a special number, equal to $$0.\overline{142857}$$. Note that there are six digits that repeat, 142857. A beautiful thing happens when we consider integer multiples of this number:
 * $$\frac{1}{7} = 0.\overline{142857}$$
 * $$\frac{2}{7} = 0.\overline{285714}$$
 * $$\frac{3}{7} = 0.\overline{428571}$$
 * $$\frac{4}{7} = 0.\overline{571428}$$
 * $$\frac{5}{7} = 0.\overline{714285}$$
 * $$\frac{6}{7} = 0.\overline{857142}$$

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 $$\frac{2}{7} = 0.\overline{285714}$$, so by moving the decimal point, we know that $$\frac{200}{7} =28.\overline{571428}$$. All that remains is to add the one from 7/7, giving us $$\frac{207}{7} =29.\overline{571428}$$.

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 of the numerator repeating to infinity:
 * $$\frac{1}{9} = 0.\overline{11}$$
 * $$\frac{2}{9} = 0.\overline{22}$$
 * $$\frac{3}{9} = 0.\overline{33}$$
 * $$\frac{4}{9} = 0.\overline{44}$$
 * $$\frac{5}{9} = 0.\overline{55}$$
 * $$\frac{6}{9} = 0.\overline{66}$$
 * $$\frac{7}{9} = 0.\overline{77}$$
 * $$\frac{8}{9} = 0.\overline{88}$$

To solve a problem such as 367/9, we reduce it to
 * $$\frac{367}{9} = \frac{300}{9} + \frac{60}{9} + \frac{7}{9}$$
 * $$\frac{367}{9} = 33.\overline{33} + 6.\overline{66} + 0.\overline{77}$$

First add $$33.\overline{33} + 6.\overline{66} = 40.0$$. Then add $$40.0 + 0.\overline{77} = 40.\overline{77}$$.

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 integer multiple repeating to infinity:
 * $$\frac{1}{11} = 0.\overline{09}$$
 * $$\frac{2}{11} = 0.\overline{18}$$
 * $$\frac{3}{11} = 0.\overline{27}$$
 * $$\frac{4}{11} = 0.\overline{36}$$
 * $$\frac{5}{11} = 0.\overline{45}$$
 * $$\frac{6}{11} = 0.\overline{54}$$
 * $$\frac{7}{11} = 0.\overline{63}$$
 * $$\frac{8}{11} = 0.\overline{72}$$
 * $$\frac{9}{11} = 0.\overline{81}$$
 * $$\frac{10}{11} = 0.\overline{90}$$
 * $$\frac{36}{11} = 3.\overline{27}$$

Cube Roots for under 1,000,000
Before this goes any further, notice the last digit for the first nine cubes. 1, 8, 27, 64, 125, 216, 343, 512, 729

Notice how the last digit for each cube contains the numbers 1-9.

For each number, break it down into two groups. The last three digits and the remaining digits at the front. For the last three digits, find the cube root whose cube ends with the same number it ends with. That will be the last digit. For the front few digits, round it down to the previous cube and find the cube root of that. That will be the first digit.

Let's take "148877" for example. Divide it into 2 groups: 148, and 877. Start with 877. The number ends with a 7 and the number 3 is the only number in which a cube ends with "7". So the last digit must be 3.

Look at 148. Round it down to the previous cube, which is 125. The cube root of 125 is 5, which is the first digit. Put it together, and 53 is the answer.

This strategy will not work on cube roots that contain decimals, although it will give an approximate.

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).

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 it's also very handy for some mental mathematics.

Other mental maths
Perhaps one of the more useful tricks to mental math is memorization. It may seem an annoyance to memorize certain math facts. However, it can tremendously speed up your calculations when you don't have to do the division or multiplication in your head. Such useful math facts to memorize include: perfect squares and cubes (especially powers of two), prime factorizations of certain numbers, and the decimal equivalents of common fractions (such as 1/7 = .1428...). For example, trying to figure out 1024/32 is much easier knowing that it's essentially the same as 2^10/2^5. Many of these are memorized simply by frequent use. Thus, the best way to get good is to 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.

Advanced Mathematics
It is possible to make good estimates of quite complex formulae. Range search is a useful technique. Aside from that, there are many other mathematical rules that you can exploit too.

Binomial Expansion
The Binomial Theorem means we can relatively easily expand an equation like:
 * $$\sqrt{1 + x} = 1 + \textstyle \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16} x^3 - \frac{5}{128} x^4 + \dots\!$$

This is very useful if $$x$$ is less than 1. In that case, the powers of $$x$$ get smaller and smaller, and we can ignore them. Even for whole numbers and Integers, this is useful as we can split the problem into a larger number and smaller number added and then raised to a power. The smaller number's powers will become less important. We can then neglect terms of the expansion that involve the smaller numbers, and thereby get a reasonable estimate. For example:
 * $$(x+y)^4 \;=\; x^4 \,+\, 4 x^3y \,+\, 6 x^2 y^2 \,+\, 4 x y^3 \,+\, y^4.$$

If the $$y$$ value is much smaller than the $$x$$, we can make a good estimate, even if we don't add the $$y^4,$$ $$4 x y^3$$, or $$6 x^2 y^2$$ terms.

The Binomial Theorem again comes to our help. Here we usually want to work out the following for compound interest:
 * $$(1+x)^n$$
 * where the original principal sum is 1, $$x$$ is the rate of interest over a short period of time (e.g. if the monthly rate is 5%, then $$x = 0.05$$), and $$n$$ is the compounding frequency over a longer period of time (for example 12, if the interest rate is the monthly rate, and the overall period is a year).

Using the theory we can roughly estimate this as
 * $$1 + n x + \frac{n (n-1)}{2} x^2$$

Although this is only a simple approximation, it works quite well for a small x.