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655-705 (Hard)|   Algebra|   Exponents|   Functions and Custom Characters|   Number Properties|                              
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MasteringGMAT
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The function f is defined for each positive three-digit integer n by 10 Aug 2023, 22:16
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solution of this questions explained in the video : https://youtu.be/PPihvpQtX6U
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suu007
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The function f is defined for each positive three-digit integer n by 13 Nov 2023, 22:57
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The simplest and quickest way to solve this is below-

Take f(v) as a value containing factors 2,3,5 which will crate 3 digit integer.
Assume f(v) as value 100 or 150 or 200 anything.

For explanation below, consider f(v) as value 200.
f(v)=200 =2^3 * 3^0 * 5^2
so, v=302

similarly f(m)=9f(v)= 2^3 * 3^2 * 5^2
so, m=322

m-v=322=302=20
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The function f is defined for each positive three-digit integer n by 27 Apr 2026, 11:44
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This question sould be more than 705+
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PGTLrowanhand
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The function f is defined for each positive three-digit integer n by 06 Jul 2026, 10:41
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Here's my video solution:

Orange08
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 \(f(m)=9*f(v)\) , then \(m-v=\) ?

(A) 8
(B) 9
(C) 18
(D) 20
(E) 80
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Spamittttttttt
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The function f is defined for each positive three-digit integer n by 23 Sep 2026, 00:50
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I thought f(n)= 2^x 3^Y 5^z - represent the 3 digits of f(n) and I was really surprised to find out that it was multiplication. Probably should have noticed how will we handle it when the powers result in double digits but yeah, it did not strike me.

Orange08
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 \(f(m)=9*f(v)\) , then \(m-v=\) ?

(A) 8
(B) 9
(C) 18
(D) 20
(E) 80
Show more
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The function f is defined for each positive three-digit integer n by 23 Sep 2026, 03:50
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Spamittttttttt
I thought f(n)= 2^x 3^Y 5^z - represent the 3 digits of f(n) and I was really surprised to find out that it was multiplication. Probably should have noticed how will we handle it when the powers result in double digits but yeah, it did not strike me.


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Yes, here x, y, and z are the three digits of n, not the digits of f(n).

For example, if n = 123, then x = 1, y = 2, and z = 3. So

f(123) = 2^1 * 3^2 * 5^3.
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