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What is the greatest prime factor of 42^2+75^2+6300 ? (A) 3

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by GMATGuruNY » Fri May 12, 2017 3:22 am
ziyuenlau wrote:What is the greatest prime factor of 42^2+75^2+6300 ?

(A) 3
(B) 7
(C) 11
(D) 13
(E) 79
The given expression looks like the red portion of the following:
(a+b)² = a² + b² + 2ab.

Test where (42 + 75)² = 42² + 75² + 6300:
(42 + 75)²

= 42² + 75² + 2*42*75

= 42² + 75² + 2*6*7*5*5*3

= 42² + 75² + 10*30*21

= 42² + 75² + 6300.
Success!

Thus:
42² + 75² + 6300 = (42+75)² = 117².
Of the five answer choices, the greatest value that divides evenly into 117 is D:
117 = 13*9.

The correct answer is D.
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Re: What is the greatest prime factor of 42^2+75^2+6300 ? (A) 3

by Scott@TargetTestPrep » Mon Apr 13, 2020 3:50 am
hazelnut01 wrote: ↑
Fri May 12, 2017 12:10 am
 What is the greatest prime factor of 42^2+75^2+6300 ?

(A) 3
(B) 7
(C) 11
(D) 13
(E) 79

OA=D
Consider the perfect square (a + b)^2, which expands to a^2 + b^2 + 2ab.

This pattern can be applied to (42 + 75)^2, which expands to 42^2 + 75^2 + 2(42)(75). Note that the third term of this expansion 2(42)(75) = 6300. Thus, we see that 42^2 + 75^2 + 6300 = (42 + 75)^2. Now, perform the addition inside the parentheses, obtaining 117^2 = (3^2 x 13)^2. We see that the greatest prime factor is 13.

Answer: D

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