vaivish wrote:hey thanks a ton...but the question posted is right.....i have checked it again...but forgot the source.....
Hi vaivish- it's not your fault, obviously, but whoever published the question didn't print it correctly. The x, y and z are supposed to be exponents. So the question should read:
The function f is defined for each positive three-digit integer n by f(n) = (2^x)(3^y)(5^z) , where x, y and z are the hundreds, tens, and units digits of n, respectively. If m and v are three-digit positive integers such that f(m)=9f(v), them m-v=?
I gave a brief solution above, but in further detail:
if m = ABC, where A, B and C are digits, and v = DEF, where D, E and F are digits, then
f(m) = 2^A * 3^B * 5^C
and
f(v) = 2^D * 3^E * 5^F
Since f(m) = 9f(v), we have:
2^A * 3^B * 5^C = 9 * 2^D * 3^E * 5^F
2^A * 3^B * 5^C = 3^2 * 2^D * 3^E * 5^F
2^A * 3^B * 5^C = 2^D * 3^(E+2) * 5^F
and by the Fundamental Theorem of Arithmetic (unique factorization into primes), if two integers are equal, they have the same prime factorizations, so the powers on the left and on the right must be equal:
A = D
B = E+2
C = F
So the number ABC is only different from DEF in the tens digit- its tens digit is two greater, so ABC is 20 larger than DEF, and m is 20 larger than v.
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