For every five hours of studying combinatorics-type questions, the average GMAT student increases their chances of being able to correctly answer a question type that is found only on the very difficult end of the GMAT spectrum. Meanwhile, the same student will have to compute hundreds of basic computations without the aid of a calculator. For students who know how to quickly do these computations, they are rewarded with extra minutes that can be spent double-checking their work and critically thinking about whether their answers make sense. As BenGMAT Franklin might say- a second saved is a second earned on the GMAT… but it doesn’t matter if those extra seconds come from being faster at doing combinatorics questions or quicker at computations. So check out these five math tricks, learn the ones that you like, and practice them daily to give yourself some extra time to finish off that 37th and final quant question.

Note: like everything else on the GMAT, being able to do something and being able to do something QUICKLY are two different tasks. If you like any of the following tricks, make sure you know it inside and out before you try using it during your test.

**1. Add or Subtract 2 or 3 Digit Numbers**

To add numbers that aren’t already a multiple of ten or one-hundred, round the number to the nearest tens or hundreds digit, add, and then add or subtract by the number you rounded off. Do the opposite when subtracting.

**Examples:**

144 + 48 = 144 + 50 – 2 = 192

1385 – 492 = 1385 – 500 + 8 = 893

**Why?**

This math trick comes down to the order of operations- adding and subtracting occur in the same step and can happen in either order. Like many other computation tricks, this one comes down to replacing one tricky computation with two simpler ones.

**2. Multiply or Divide by 5**

To multiply a number by 5, divide by 2 and then multiply by 10. To divide a number by 5, divide by 10 and then multiply by 2.

**Example:**

82 Ã— 5 = 82 Ã· 2 x 10 = 410

**Why?**

This math trick comes down to the order of operations- multiplying and dividing occur in the same step and can happen in either order. But instead of doing the (somewhat) difficult task of multiplying by 5, do the easier task of multiplying by the fraction 10/2. And since you can do this in either order, you can start by dividing a number by 2 or multiplying the number by 10. Starting with division is usually easier when you start with an even number (34 Ã— 5 = 17 x 10 = 170) while starting with multiplication is easier when beginning with a non-integer (6.4 Ã— 5 = 64 Ã· 2 = 32). And instead of thinking about dividing by 5, think about multiplying by 2/10 (455 Ã· 5 = 45.5 Ã— 2 = 91).

**3. Multiply Numbers Between 11 & 19**

To multiply two numbers that are between 11 and 19, add the ones digit of one number to the other number, multiply by 10, and then add the product of the ones digits.

**Example:**

14 Ã— 13 = (17 Ã— 10) + (4 Ã— 3) = 182

**Why?**

In the standard way that most American-students are taught to multiply numbers, you set up two numbers on top of one another like this:

14

Ã— 13

42

140

182

This leads to a problem where you multiply 3 by 14, then multiply 10 by 14 and add the two products together. But you can rearrange this problem further to say you want to multiply 3 by 4, 3 by 10, and 10 by 14. Because you are multiplying both 3 and 14 by the same factor of 10 (which only happens when both numbers are between 11-19), you can combine this into one step. So instead of doing one tricky computation (3 Ã— 14) and two easy ones (10 Ã— 14 and 42 + 140), you make four easy computations (14 + 3, 17 Ã— 10, 4 Ã— 3, and 170 + 12).

**4. Square Any Number Between 11 & 99**

To square any number n, first find the nearest multiple of 10 and find out how much you would have to add or subtract (k) to get to that number. Then do the opposite function (addition or subtraction) to get two numbers that average out to n (i.e. n + k and n – k). Multiply those two numbers and add the square of k.

**Examples:**

23^{2} = (26 x 20) + 3^{2} = 529

97^{2} = (100 x 94) + 3^{2} = 9409

**Why?**

a^{2} – b^{2} = (a+b)(a-b)

a^{2} = (a+b)(a-b) + b^{2}

If this special product doesn’t look familiar to you, write it down right now and memorize it because there are a plethora of GMAT questions that test you on this very concept. But for this special trick, you are (once again) trading a difficult calculation (23 x 23) for a few simpler ones.

23^{2} – 3^{2} = (23+3)(23-3)

23^{2} – 3^{2} = (26)(20)

23^{2} = (26)(20) + 3^{2}

23^{2} = 529

Since multiplying by multiples of ten are usually easier than non-multiples of ten, you find the nearest multiple of ten. While this may be very confusing at first, it’s a neat trick if you can get quick with it and is especially helpful when squaring numbers ending in five, since you will always add 25 to the lower and higher multiple of 10:

45^{2} = (40 x 50) + 5^{2} = 2025

65^{2} = (60 x 70) + 5^{2} = 4225

**5. Estimate Root 2 & Root 3**

âˆš2 â‰ˆ 1.4

âˆš3 â‰ˆ 1.7

**Example:**

The length of some side of a figure is about equal to 13 and you are down to the following three options:

(A) 9 âˆš2

(B) 9 / âˆš2

(C) 15 * (1 – 1/âˆš2)

By estimating âˆš2 â‰ˆ 1.4 or even 3/2, you could quickly recognize that only answer (A) could be correct in this problem.

**Why?**

Because your calculator says so! But rather than trying to remember these two roots on test day, remember your two favorite holidays in the second and third months of the year- Valentine’s Day & St. Patrick’s Day, 2/14 and 3/17. Root of 2 is about 1.4 and Root of 3 is about 1.7. The above example is adapted from an old Official Guide problem and is a reminder of how the GMAT’s wrong answer choices aren’t there because they are close to correct value, but because they are some incorrect computational jumble of the numbers given in the problem. If you can estimate in this problem, you can find four incorrect answer choices and that’s just as good as finding one correct answer.

Squaring a number that ends in a 5 :

To square a number ending in a 5, try the following:

35 squared = 3 x 4 = 12 – Hence the answer is 1225 (always add a 25 at the end)

65 squared = 6 x 7 = 42 – Hence the answer is 4225

75 squared = 7 x 8 = 56 – Hence the answer is 5625

105 squared = 10 x 11 = 110 – Hence the answer is 11025

{5x[n+(n+1)]}^2 = n(n+1)x100+25.

Fantastic, amazing post!! A true evidence that Math is reinventing itself every day, I am impressed!

These are simply awesome tricks that I have ever seen in any book so far !! Thanks a lot Joe !

Hi Salmon – this is great! But can you explain where the 7 comes from in 65 squared, the 8 in 75 square, etc?

It’s always first number + 1 i.e. sqr 35 is 3xfirst number +1 i.e. 4 and then add 25

Yes, it is always 1 more than the number formed by the digits apart from the units digit. So for 85 squared, u use 8×9 =72.

kcb… its like square of 45.. jssh take first digit ‘4’ away…add both digit ‘4’ n ‘5’ … so it will be ‘9’ n deduct 4 from 9… its became 5 after deducting… nw v hv dat initial 4 n this 5… multiply it,got ’20’…..’25’as defined so became 2025 as answer….

deducting 4 is crucial from every addition

squre of 65 is ->

first num.x first num+1 and add 25

Eg:65

6x(6+1) ie; 6×7 and add 25=3025

Try this for multiplying a number by 11.

spread the original number, and put the sum of the digits in the middle. Hence 24 x 11 = 2 (2+4) 4 = 264

43 x 11 = 473

51 x11 = 561

Never saw this one before, but that’s a great trick. Also, as I’m sure Salman has already, notice how once the sum gets larger than 9, simply add the tens digit to the earlier number:

58 x 11 = 5 (5+8) 8 = 5 (13) 8 = 638

Anyone have other good tricks?

Hi Salman and Joe,

How would you use the trick of multiplying 3 digit or 4 digit numbers by 11?

how about multi plyong 2 and 3 digits

excuse me it’s multiplying

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For multiplying a larger digit by 11, I simply write down the number and add a 0 at the end, then write the same number below, and do a quick addition. Even though this technique is not ground breaking, it allows me to reduce my thinking time to zero

Hence, for example, 245 x 11 = 2450

+245

=2695 Ans

The formatting sort of distorted the presentation. Ideally you would want the bigger number right above the smaller number to allow a quick addition .

Let me try again:

245 x 11 =

2450

+245

2695 – Ans

a damn common sense… 245(10+1) its 2450+245=2695

For multiplying a larger digit by 11, I simply write down the number and add a 0 at the end, then write the same number below, and do a quick addition. Even though this technique is not ground breaking, it allows me to reduce my thinking time to zero

Hence, for example, 245 x 11 =

2450

+245

=2695 Ans

Really, you are great!!!!!!!! Thanks

There is also another trick for multiplying 3 or more digits by 11.

For ex to multiply 13524 by 11 you can do the following:

Step 1: Write 1 and 4 as your first and last number respectively.

Step 2: start from the left most number and make pairs of each consecutive terms (which you later sum). The first pair will be 2 and 4 (for a sum of 6). The next will be 5 and 2 (sum of 7) followed by 3 and 5 (sum of 8) and the last pair will be 1 and 3 (sum of 4).

Step 3. write them in their respective orders starting from the left (keeping in mind step 1).

hence you get 148764

It’s pretty easy once you write it down.

p.s. always remember you start writing digits from the left since you will need to carry to the next number (sum) if the sum is 10 or more.

Great Tips….