What is the greatest prime factor of 2^10*5^4 – 2^13*5^2 + 2^14?

time consuming, but definitely not a 700-level question.
we can factor 2^10 first, and we remain with 5^2(5^2 - 2^3)+2^4

25-8 = 17, so we have 25*17 + 16, which is 441. I did not start doing the prime factorization, since it is more time consuming than trying to divide by each option choice.
it is definitely not divisible by 11, so D is out.
Divide by 13, well, 13*3 = 39, and we are left with 51, which is not divisible by 13, E out.
Try by 7, works, so definitely the GCF is 7.

Nice Question
Here is what i did

N=2^10[625-200+16]
N=2^10*441
N=2^10*3^2*7^2
Hence 7 is the highest prime factor.
Hence C
Its just a time consuming Question.

Similar questions:
what-is-the-greatest-prime-factor-of-158991.html
what-is-the-greatest-prime-factor-of-104757.html
what-is-the-greatest-prime-factor-of-70126.html

Answer is C

On solving further , we arrive at 2^10 * 441 and that follows 2^10 * 7^2*3^2 so the greatest prime factor is "7"

Basically you have to simplify the expression.
Take 2^10 out of the expression; you get 2^10*441=2^10*21^2
That gives 7 as the greatest prime factor.

Bunuel shouldnt initiall question look like this \((2^{10}*5^4) - (2^{13}*5^2)+2^{14}\)

\(2^{10}*5^2(5^2-2^3)+2^{14}\)

\(2^{10}*25(25-8)+2^{14}\)

\(2^{10}*25(17)+2^{14}\)

\(2^{10}(425)+2^{14}\)

\(2^{10}(425+16)\)

\(2^{10} (441)\)

Prime factors of 441 are 3 and 7

So, the greatest prime factor is 7

Looking at the expression, we can say for sure that 2 is a factor and 5 is not.
Beyond this, if we have to find other factors, we will have to simplify the expression.

\(2^{10}*5^4 - 2^{13}*5^2 + 2^{14}\)
We have \(2^{10}\) common in all terms of the expression, so we can factor that out.

\(2^{10}(5^4 - 2^{3}*5^2 + 2^{4})=2^{10}(625-200+16)=2^{10}*441\)

\(2^{10}*21^2=2^{10}*3^2*7^2\)

Greatest prime factor is 7.


C

it just requires slight simplification
=>2^10[625-200+16]
=>2^10*441
=>2^10*3^2*7^2

Therefore IMO C

If you look closely this is a Square of a Difference Quadratic Template:

(X - Y)^2 = (X)^2 - (2)(X)(Y) + (Y)^2


X = (2)^5 * (5)^2

Y = (2)^7

[ (2)^5 * (5)^2 - (2)^7 ] ^2 =


[ (2)^5 * (5)^2 - (2)^7 ] * [ (2)^5 * (5)^2 - (2)^7 ] =

(2)^5 * (5)^2 * (2)^5 * (5)^2 - (2) * (2)^5 * (5)^2 * (2)^7 + (2)^7 * (2)^7 =


(2)^10 * (5)^4 - (2)^13 * (5)^2 + (2)^14


Thus, using the Square of a Difference Quadratic Template, we can focus on one of the Factors (other one will be identical)


[ (2)^5 * (5)^2 - (2)^7 ]

-take common factor of (2)^5

(2)^5 * [ (5)^2 - (2)^2 ]

If there is a larger prime factor it will come from the expression within the brackets

(5)^2 - (2)^2 =

25 - 4 = 21 = (7) (3)


(C) 7 is the largest prime factor

Posted from my mobile device

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.