Manhattan GMAT Challenge Problem of the Week - 14 Sept 2011

by Manhattan Prep, Sep 14, 2011

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Question

If the quantity [pmath]5^2[/pmath]+ [pmath]5^4[/pmath] + [pmath]5^6[/pmath] is written as (a + b)(a b), in which both a and b are integers, which of the following could be the value of b?

A. 5

B. 10

C. 15

D. 20

E. 25

Answer

One first natural move is to factor [pmath]5^2[/pmath] out of the numeric expression:

[pmath]5^2[/pmath] + [pmath]5^4[/pmath] + [pmath]5^6[/pmath] = [pmath]5^2[/pmath](1 + [pmath]5^2[/pmath] + [pmath]5^4[/pmath])

Meanwhile, (a + b)(a b) becomes [pmath]a^2[/pmath] [pmath]b^2[/pmath].

So, somehow we have to turn [pmath]5^2[/pmath](1 + [pmath]5^2[/pmath] + [pmath]5^4[/pmath]) into a difference of squares. It must come down to the stuff in the parentheses.

Flip around those terms, so that we have the larger powers first: [pmath]5^4[/pmath] + [pmath]5^2[/pmath] + 1.

This should look almost like a special product: [pmath](x + y)^2[/pmath] = [pmath]x^2[/pmath] + [pmath]2xy[/pmath] + [pmath]y^2[/pmath], if x = [pmath]5^2[/pmath] and y = 1. In fact, write that out:

[pmath](5^2+1)^2[/pmath] = [pmath]5^4[/pmath] + 2([pmath]5^2[/pmath]) + 1

The expression we actually have, [pmath]5^4[/pmath] + [pmath]5^2[/pmath] + 1, is very close. Add [pmath]5^2[/pmath] and subtract it as well:

[pmath]5^4[/pmath] + [pmath]5^2[/pmath] + 1

= [pmath]5^4[/pmath] + [pmath]5^2[/pmath] + 1 + [pmath]5^2[/pmath] [pmath]5^2[/pmath]

= ([pmath]5^4[/pmath] + [pmath]5^2[/pmath] + 1 + [pmath]5^2[/pmath]) [pmath]5^2[/pmath]

= ([pmath]5^4[/pmath] + 2([pmath]5^2[/pmath]) + 1) [pmath]5^2[/pmath]

= [pmath](5^2 + 1)^2[/pmath] [pmath]5^2[/pmath]

= [pmath]26^2[/pmath] - [pmath]5^2[/pmath]

Almost there. Remember, we had [pmath]5^2[/pmath] as well:

[pmath]5^2[/pmath] + [pmath]5^4[/pmath] + [pmath]5^6[/pmath] = [pmath]5^2[/pmath](1 + [pmath]5^2[/pmath] + [pmath]5^4[/pmath]) = [pmath]5^2[/pmath]([pmath]26^2[/pmath] - [pmath]5^2[/pmath])

= [pmath](26*5)^2[/pmath] [pmath](5*5)^2[/pmath]

= [pmath]130^2[/pmath] [pmath]25^2[/pmath].

Oher differences of squares can be equal to the original expression, but this is the only one that fits an answer choice: (E) 25.

There are other ways to solve this problem as well, such as backsolving.

[pmath]5^2[/pmath] + [pmath]5^4[/pmath] + [pmath]5^6[/pmath] = [pmath]a^2[/pmath] [pmath]b^2[/pmath]

[pmath]5^2[/pmath] + [pmath]5^4[/pmath] + [pmath]5^6[/pmath] + [pmath]b^2[/pmath] = [pmath]a^2[/pmath]

Try different bs from the answer choices, and see which one, when added to the original expression, gives you a perfect square.

Testing (E):

[pmath]5^2 [/pmath]+ [pmath]5^4[/pmath] + [pmath]5^6[/pmath] + [pmath]25^2[/pmath]

= [pmath]5^2[/pmath] + [pmath]5^4[/pmath] + [pmath]5^6[/pmath] + [pmath]5^4[/pmath]

= [pmath]5^2[/pmath] + 2[pmath]5^4[/pmath] + [pmath]5^6[/pmath]

= [pmath](5 + 5^3)^2[/pmath]

This is the only answer that fits.

The correct answer is E.

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