functions

[vishal chugh]

The function f is defined for each positive three-digit integer n by f(n) = 2^x3^y5^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=?
(A) 8
(B) 9
(C) 18
(D) 20
(E) 8

[pemdas]

2^x3^y5^z=9*(2^X3^Y5^Z)=2^X3^(Y+2)5^Z,
it's clear that the difference between two numbers is in the ten's unit by 2
since unit's digits are the same and we have difference only in ten's digit, the difference in unit's digit=0 and we look for difference in ten's digit, which is d=20

The function f is defined for each positive three-digit integer n by f(n) = 2^x3^y5^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=?
(A) 8
(B) 9
(C) 18
(D) 20
(E) 8

[GMATGuruNY]

The function f is defined for each positive three-digit integer n by f(n) = 2^x3^y5^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=?
(A) 8
(B) 9
(C) 18
(D) 20
(E) 8

Let m = 120.
f(120) = (2^1)(3^2)(5^0) = 18.

Since f(m) = 9f(v):
18 = 9f(v).
f(v) = 2.

If f(v) = 2, then v=100:
f(100) = (2^1)(3^0)(5^0) = 2.

Thus, m-v = 120-100 = 20.

The correct answer is D.

[ollapodrida]

The function f is defined for each positive three-digit integer n by f(n) = 2^x3^y5^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=?
(A) 8
(B) 9
(C) 18
(D) 20
(E) 8

Let m = 120.
f(120) = (2^1)(3^2)(5^0) = 18.

Since f(m) = 9f(v):
18 = 9f(v).
f(v) = 2.

If f(v) = 2, then v=100:
f(100) = (2^1)(3^0)(5^0) = 2.

Thus, m-v = 120-100 = 20.

The correct answer is D.

What motivated your choice of m=120 in this problem?

[ollapodrida]

2^x3^y5^z=9*(2^X3^Y5^Z)=2^X3^(Y+2)5^Z,
it's clear that the difference between two numbers is in the ten's unit by 2
since unit's digits are the same and we have difference only in ten's digit, the difference in unit's digit=0 and we look for difference in ten's digit, which is d=20

I don't quite follow your logic. Can you add a little more detail or break things down to make your logic more clear here?

[shikh]

can u plz explain it more clearly

[shankar.ashwin]

You are given f(m) = 9f(v)

Assume minimum value of 'v' = 1 (001)

Therefore f(v) = 2^0 * 3^0 * 5^1 = 5

Since, f(m) = 9* f(v) = 9*5 = 2^0 * 3^2 * 5^1.

THe powers here correspond to x,y,z respectively (or) m = 021

m-v = 21-1 = 20 D

[GMATGuruNY]

The function f is defined for each positive three-digit integer n by f(n) = 2^x3^y5^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=?
(A) 8
(B) 9
(C) 18
(D) 20
(E) 8

Let m = 120.
f(120) = (2^1)(3^2)(5^0) = 18.

Since f(m) = 9f(v):
18 = 9f(v).
f(v) = 2.

If f(v) = 2, then v=100:
f(100) = (2^1)(3^0)(5^0) = 2.

Thus, m-v = 120-100 = 20.

The correct answer is D.

What motivated your choice of m=120 in this problem?

Since f(m) = 9f(v), f(m) is a multiple of 9.
If three-digit m = 120, then f(120) = (2^1)(3^2)(5^0) = 18, which is a multiple of 9.

[adamsmith2009]

Isn't the question asking (2^x)(3^y)(5^z)?

The way you wrote it: f(n) = 2^x3^y5^z it looks like everything is two is being raised with all of the powers or is it just me?

[tuanquang269]

Thank guru for explanation

[nskandan]

I went through a more linear route
m = xyz, v = abc

(2^x)(3^y)(4^z) = 9. (2^a)(3^b)(4^c)
(2^(x-a))(3^(y-b))(4^(z-c)) = 9

so y-b = 2 and x-a & z-c has to be 0

hence m-v = xyz-abc = 20