Manhattan GMAT Challenge Problem of the Week – 12 November 2012

by on November 12th, 2012

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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:

2^41^4 = 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:

3^42^4 = 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^36n^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.

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2 comments

  • 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).

    • 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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