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Manhattan GMAT Challenge Problem of the Week – 12 November 2012

by Manhattan GMAT on November 12th, 2012
2 comments
Click here to read more articles from Manhattan GMAT and to learn more about Manhattan GMAT's classes.
Posted in
  • Arithmetic
  • Data Sufficiency
  • GMAT Math
  • Integer Properties
  • Powers and Roots

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

Question

The positive difference of the fourth powers of two consecutive positive integers must be divisible by

A. one less than twice the larger integer
B. one more than twice the larger integer
C. one less than four times the larger integer
D. one more than four times the larger integer
E. one more than eight times the larger integer

Answer

You could solve this problem algebraically, or you could solve by picking numbers. Let’s try the latter method first, as it winds up being faster and easier mentally.

Write the fourth powers of the first three positive integers:
1^4 = 1
2^4 = 16
3^4 = 81

Now, for the first two fourth powers, the “positive difference of the fourth powers of two consecutive positive integers” would be this:

2^4 – 1^4 = 16 – 1 = 15

15 is divisible by 3 and by 5 (as well as by 1 and by 15). Since the “larger integer” is 2, eliminate answers:

(A) one less than twice the larger integer = 2×2 – 1 = 3 = fine
(B) one more than twice the larger integer = 2×2 + 1 = 5 = fine
(C) one less than four times the larger integer = 4×2 – 1 = 7 = wrong
(D) one more than four times the larger integer = 4×2 + 1 = 9 = wrong
(E) one more than eight times the larger integer = 8×2 + 1 = 17 = wrong

You’re left with (A) and (B). Now use 2 and 3 as your consecutive integers:

3^4 – 2^4 = 81 – 16 = 65

65 is divisible by 5 and by 13 (as well as by 1 and by 65). Test (A) and (B):

(A) one less than twice the larger integer = 2×3 – 1 = 5 = fine

(B) one more than twice the larger integer = 2×3 + 1 = 7 = wrong

The answer must be (A).

As for the algebraic proof, here it is:

“The positive difference of the fourth powers of two consecutive positive integers”

= n^4 – (n-1)^4

If you expand the (n-1)^4 term and do the subtraction, you get 4n^3 – 6n^2 + 4n – 1, which is hard to factor further. Here’s the trick: you can treat n^4 – (n-1)^4 as the difference of squares, since every fourth power is a square. For instance n^4 = (n^2)^2. So you get this:
n^4 – (n-1)^4 = [n^2 + (n-1)^2] x [n^2 –  (n-1)^2]

Now expand the right side:

n^2 + (n-1)^2 = n^2 + (n^2 – 2n + 1) = 2n^2 – 2n + 1

n^2 – (n-1)^2 = n^2 – (n^2 – 2n + 1) = 2n – 1

So you get:

n^4 – (n-1)^4 = (2n^2 – 2n + 1) (2n – 1)

Thus, as long as n is an integer, the expression on the left is always divisible by 2n – 1, which corresponds to choice A.

The correct answer is A.

Special Announcement: If you want to win prizes for answering our Challenge Problems, try entering our Challenge Problem Showdown. Each week, we draw a winner from all the correct answers. The winner receives a number of our our Strategy Guides. The more people enter, the better the prize. Provided the winner gives consent, we will post his or her name on our Facebook page.

If you liked this article, let Manhattan GMAT know by clicking Like.

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2 comments

  • Jinwoo on November 12th, 2012 at 5:31 am

    Be careful of term "larger integer" which should be "n".
    If you starts as (n+1)^4-(n)^4, the fomular will mislead you (B).

    Reply to this comment
    • Rahul Sehgal on November 18th, 2012 at 11:48 am

      Great point Jinwoo. This is what I did when I was trying to solve this question. But, I am still not sure why it did not work if we took the numbers as n+1 and n ; considering n+1 as the greater consecutive integer. Any thoughts please.

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