functions

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vishal chugh: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Fri Nov 18, 2011 11:28 pm; Location: Ludhiana ,Punjab, India; Thanked: 1 times

by vishal chugh » Sat Nov 19, 2011 12:11 am

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

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Source: Beat The GMAT — Problem Solving | Sat Nov 19, 2011 12:11 am

pemdas: Legendary Member; Posts: 1084; Joined: Fri Apr 15, 2011 2:33 pmFollowed by: 21 members; Thanked: 158 times

by pemdas » Sat Nov 19, 2011 12:21 am

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

vishal chugh wrote: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

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Nov 19, 2011 5:42 am

vishal chugh wrote: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.

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ollapodrida: Senior | Next Rank: 100 Posts; Posts: 37; Joined: Thu Oct 06, 2011 2:49 pm; Thanked: 4 times

by ollapodrida » Sun Nov 20, 2011 6:54 pm

GMATGuruNY wrote:
vishal chugh wrote: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?

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ollapodrida: Senior | Next Rank: 100 Posts; Posts: 37; Joined: Thu Oct 06, 2011 2:49 pm; Thanked: 4 times

by ollapodrida » Sun Nov 20, 2011 7:01 pm

pemdas wrote: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?

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shikh: Senior | Next Rank: 100 Posts; Posts: 48; Joined: Sun Jul 17, 2011 12:04 am

by shikh » Sun Nov 20, 2011 11:08 pm

can u plz explain it more clearly

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shankar.ashwin: Legendary Member; Posts: 966; Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members; Thanked: 230 times

by shankar.ashwin » Sun Nov 20, 2011 11:43 pm

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

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Mon Nov 21, 2011 4:22 am

ollapodrida wrote:
GMATGuruNY wrote:
vishal chugh wrote: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.

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adamsmith2009: Senior | Next Rank: 100 Posts; Posts: 36; Joined: Fri Mar 27, 2009 6:53 pm

by adamsmith2009 » Tue Nov 22, 2011 10:53 am

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?

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tuanquang269: Master | Next Rank: 500 Posts; Posts: 296; Joined: Sun May 29, 2011 5:10 amFollowed by: 5 members; Location: Vietnam; Thanked: 10 times

by tuanquang269 » Tue Nov 22, 2011 8:32 pm

Thank guru for explanation

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nskandan: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Mon Oct 10, 2011 3:06 am; Location: UK; Thanked: 1 times

by nskandan » Wed Nov 23, 2011 8:52 am

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

NSK

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