BrentGMATPrepNow wrote:
\(K = 10^{26} + 2^{26}\), and \(K\) is a multiple of \(2^n\) but NOT a multiple of \(2^{n+1}\). If n is a positive integer, what is the value of \(n+1\)?
A) 25
B) 26
C) 27
D) 28
E) 29
IMPORTANT: later in my solution, we need to recognize that 5^b will end in 25 for all integer values of b greater than 0. For example, 5^2 =
255^3 = 1
255^4 = 6
255^5 = 31
25 5^6 = XXX
25 etc....
So......
K = 10^26 + 2^26
= 2^26(5^26 + 1)
= 2^26(XXXX
25 + 1)
[aside: XXXX25 denotes some number ending in 25]= 2^26(XXXX
26)
= 2^26[
2(XXXX
3)]
[Since XXX26 is EVEN, we can factor out a 2]=
2^27[XXXX3] Since XXXX
3 is an ODD number, we can't factor out any more 2's.
This means K is a multiple of
2^27, but K is NOT a multiple 2^28.
In other words,
n = 27What is the value of n + 1? Since n = 27, we can conclude that n + 1 = 27 + 1 = 28
Answer: D
Cheers,
Brent
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