Bunuel wrote:
gmatcracker2017 wrote:
Bunuel wrote:
5. If \(x=23^2*25^4*27^6*29^8\) and is a multiple of \(26^n\),where \(n\) is a non-negative integer, then what is the value of \(n^{26}-26^n\)?
A. -26
B. -25
C. -1
D. 0
E.1
\(23^2*25^4*27^6*29^8=odd*odd*odd*odd=odd\) so\(x\) is an odd number. The only way it to be a multiple of\(26^n\) (even number in integer power) is when\(n=0\) , in this case\(26^n=26^0=1\) and 1 is a factor of every integer. Thus\(n=0\) -->\(n^{26}-26^n=0^{26}-26^0=0-1=-1\) .Must know for the GMAT :\(a^0=1\) , for\(a\neq{0}\) - any nonzero number to the power of 0
A. -26
B. -25
C. -1
D. 0
E.1
\(23^2*25^4*27^6*29^8=odd*odd*odd*odd=odd\) so\(x\) is an odd number. The only way it to be a multiple of\(26^n\) (even number in integer power) is when\(n=0\) , in this case\(26^n=26^0=1\) and 1 is a factor of every integer. Thus\(n=0\) -->\(n^{26}-26^n=0^{26}-26^0=0-1=-1\) .Must know for the GMAT :\(a^0=1\) , for\(a\neq{0}\) - any nonzero number to the power of 0
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