GMAT prep (from the web)

[sudaif]

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), then m-v = ?
A. 8
B. 9
C. 18
D. 20
E. 80

Not sure how to begin solving this problem, because we don't know about m's or v's digits.

[adiagr]

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), then m-v = ?
A. 8
B. 9
C. 18
D. 20
E. 80

Not sure how to begin solving this problem, because we don't know about m's or v's digits.

It is given that f(m) = 9 f(v); Now 9 = 3^2.

say
v = (2^x).(3^y).(5^z)

Then

m = (9).(3^y).(5^z).(2^x) = (3^2).(3^y).(5^z).(2^x)
m= (3^y+2).(5^z).(2^x)

This means that for m and v, Hundred's digit and unit's digit are same. However, Tens digit is y in case of v and (y+2) in case of m.


For example:

say

m=135 [Hundred's digit is 3]
v=115 [Unit's digit is 1]

m-v=20

so always m-v will be 20. Option (D) is the answer.

Aditya

[mschwrtz]

Very nice, and exactly right. The key here is that the only way that f(m) can equal 9 times f(v) is to have two additional 3s among its prime factors. No product of 2s and 5s is going to get you 9.

The only way that the prime factorization for f(m) can have two more 3s than the prime factorization for f(v) is if m has a tens digit 2 greater than v.

[sudaif]

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), then m-v = ?
A. 8
B. 9
C. 18
D. 20
E. 80

Not sure how to begin solving this problem, because we don't know about m's or v's digits.

It is given that f(m) = 9 f(v); Now 9 = 3^2.

say
v = (2^x).(3^y).(5^z)

Then

m = (9).(3^y).(5^z).(2^x) = (3^2).(3^y).(5^z).(2^x)
m= (3^y+2).(5^z).(2^x)

This means that for m and v, Hundred's digit and unit's digit are same. However, Tens digit is y in case of v and (y+2) in case of m.


For example:

say

m=135 [Hundred's digit is 3]
v=115 [Unit's digit is 1]

m-v=20

so always m-v will be 20. Option (D) is the answer.

Aditya

makes sense. good job!

[RonPurewal]

nice work.