Manhattan GMAT Challenge Problem of the Week – 12 July 2010

by Manhattan Prep, Jul 13, 2010

Here is a new Challenge Problem! If you want to win prizes, try entering our Challenge Problem Showdown. The more people enter our challenge, the better the prizes.

As always, the problem and solution below were written by one of our fantastic instructors. Each challenge problem represents a 700+ level question. If you are up for the challenge, however, set your timer for 2 mins and go!

Question

What is the 99th digit after the decimal point in the decimal expansion of 2/9 + 3/11?

A. 1

B. 2

C. 4

D. 7

E. 9

Answer

There are two ways to approach this problem. Well go through them in turn.

(1) Figure out the decimal expansions of the two fractions separately (using long division), then add.

2/9 = 0.2222

3/11 = 0.2727

Thus, the sum is 0.4949 The first digit is a 4, as is the third digit, the fifth digit, and every digit after that in an odd-numbered position. (The digits in even-numbered positions are all 9s.) Thus, the 99th digit must also be a 4.

(2) Add the fractions together, then figure out the decimal expansion.

2/9 + 3/11 = 22/99 + 27/99 = 49/99

Now, you can either figure out that the decimal expansion of 49/99 is 0.4949, or you can simply know a shortcut: any two-digit number divided by 99 becomes a decimal with that two-digit number repeating. For instance, 17/99 = 0.1717, 91/99 = 0.9191, and so on. (This pattern generalizes to any number of digits, as long as the denominator is composed of only 9s. For instance, 125/999 = 0.125125)

The final analysis is the same: every digit in an odd-numbered position is a 4, so the 99th digit after the decimal point is a 4 as well.

The correct answer is (C).

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