Therefore, if you are aiming to tackle even the 160-170 questions on the Q section, you need to have some crafty mental math tricks up your sleeve. One of the most powerful involves the clever use of a famous algebra formula: the difference of two squares formula. See this page — "Quant: Difference of Two Squares" for uses of that formula in general problem solving. Here we will focus on factoring.

## Factoring Big Numbers

In general, factoring a big number can be time-consuming without a calculator. The QUANT might ask a question in which you need to know the factors, or the prime factorization, of a large number. See this page — "Math: Factors" — for more on prime factorizations.

First of all, notice how easy it is to square the multiples of 10. If you can square the numbers from 1 to 10, you can square the multiples of 10 from 10 to 100.

10^2 = 100

20^2 = 400

30^2 = 900

40^2 = 1600

50^2 = 2500

60^2 = 3600

70^2 = 4900

80^2 = 6400

90^2 = 8100

100^2 = 10000

Those should all be recognizable as nice round perfect squares. Now, suppose you are in a situation in which you have to factor, say, 1591, or 3551, or 8099. Notice, all of those are a perfect square less than one of these multiples of ten squared.

1591 = 1600 - 9 = (40)^2 - (3)^2 = (40 + 3)(40 - 3) = 43 \times 37

3551 = 3600 - 49 = (60)^2 - (7)^2 = (60 + 7)(60 - 7) = 67 \times 53

8099 = 8100 - 1 = (90)^2 - (1)^2 = (90 + 1)(90 - 1) = 91 \times 89 = 7 \times 13 \times 89

In general, the QUANT is not going to put you in a situation in which you have to find the prime factorization of a general four digit number. If this situation does arise, you can bet there’s an enormously simplifying trick available, and factoring via the difference of two squares is an awfully likely candidate for that trick.

## Factoring Decimals

Just as the difference of two square can simplify factoring big numbers, it can also simplify factoring decimals. To demonstrate this, I am going to show the solution to a flamboyantly recondite question.

Q1. \dfrac{0.99999999}{1.0001} - \dfrac{0.99999991}{1.0003}=

(A) 10^{-8}

(B) 3(10^{-8})

(C) 3(10^{-4})

(D) 2(10^{-4})

(E) 10^{-4}

Notice that 1.0001 = 1 + \Big(10^{-4}\Big), and 1.0003 = 1 + 3\Big(10^{-4}\Big).

Now, notice that both numerators simplify via difference of two squares formula.

0.99999999 = 1-10^{-8} = 1^2-\Big(10^{-4}\Big)^2 =\Big(1+10^{-4}\Big)\Big(1-10^{-4}\Big)

0.99999991 = 1-9\Big(10^{-8}\Big) = 1^2-\Big[3\Big(10^{-4}\Big)\Big]^2 = \Big[1+3\Big(10^{-4}\Big)\Big] \times \Big[1-3\Big(10^{-4}\Big)\Big]

Thus, the two fractions become

\dfrac{0.99999999}{1.0001} = \Big(1+10^{-4}\Big)\dfrac{\Big(1-10^{-4}\Big)}{\Big(1+10^{-4}\Big)} = 1 - 10^{-4}

\dfrac{0.99999991}{1.0003} =\Big[1+3\Big(10^{-4}\Big)\Big]\dfrac{\Big[1-3\Big(10^{-4}\Big)\Big]}{\Big[1+3\Big(10^{-4}\Big)\Big]} = 1 - 3\Big(10^{-4}\Big)

The difference:

\Big(1 - 10^{-4}\Big) - \Big[1 - 3\Big(10^{-4}\Big)\Big] = 3\Big(10^{-4}\Big)-10^{-4} = 2\Big(10^{-4}\Big)

Q2. \dfrac{49^2 - 35^2}{14} =

(A) 74

(B) 76

(C) 78

(D) 79

(E) 84

Crunching these numbers is cumbersome and, fortunately, unnecessary. Our numerator is a difference of squares, meaning we can break it up by factoring.

This gives us the much simpler \dfrac{(49 + 35)(49 - 35)}{14}.

This simplifies to \dfrac{(84)(14)}{14}.

We can cancel the 14s, which leaves us with 84