love it when someone posts such effective and minimalistic explanations! Kudos to you

Hi All,

While this prompt is 'complex-looking', it's based on some relatively simple concepts (once you unravel the information that you've been given).

To start, we're told that each DIGIT of a positive 3-digit number is defined by...
f(N) = (2^X)(3^Y)(5^Z) where X, Y and Z are the hundreds, tens and units digits of N.

For example, IF... N = 410, then f(410) = (2^4)(3^1)(5^0) = (16)(3)(1) = 48

We're told that M and V are 3-digit numbers such that f(M) = 9(f(V)). We're asked for the value of M-V.

Given how the function multiplies 'powers of primes', for one 3-digit number to be 9 TIMES another, the larger number MUST have "two more 3s" than the smaller number. For that to occur, the tens digit of the larger number must be 2 greater than the tens digit of the smaller number. The other two digits (hundreds and units) must be the SAME.

For example: 240 and 220
f(240) = (2^2)(3^4)(5^0) = (4)(81)(1) = 324
f(220) = (2^2)(3^2)(5^0) = (4)(9)(1) = 36
324 = 9 times 36

Thus, the difference between the M and V comes down to the tens digits - and while there are several potential options here - they ALL differ by 20.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
_________________

This Q can be easily solved within 1 minute by the substitution-method.
As soon as i read "positive three-digit integer", i substituted v as 100 (considering m to be greater, i did not substitute m as 100)
f(v)= 2^1*3^0*5^0 = 2
Now, f(m)=9*f(v) means f(m)=9*2=18
Then prime factorizing 18 we get 2^1*3^2 which is equal to 2^1*3^2*5^0 which is equal to f(120)
Therefore f(120)=f(m), and m-v = 120-100 = 20
_________________

Hi there

I’m not sure I understand the underlying logic. Could you explain more please? Thank you!

Posted from my mobile device

This might be considered a bit of a lucky guess, but I just recognized that with bases 2 and 5, I can always have 100 by squaring both respective numbers. Then I recognized that to have f(m)=9f(v) I can have arrange the powers of 2,3,5 such that the function f(n) would return 100, and also return 900.

To have f(n)=f(m)=100, I set x=2, y=0 and z=2 resulting in a 3-digit number of 202
To have f(n)=f(v)=900, I set x=2, y=2, and z=2 resulting in a 3-digit numer of 222

Therefore, I subtract 222 from 202 and end up with 20, and hence answer choice (D)
_________________

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 \(m-v=\) ?

Let f(v) = 2^x3^y5^z
f(m) = 9*f(v) = 2^x3^{y+2}5^z

Tenth digit of m is 2 greater than tenth digit of v
m - v = 2*10 = 20

IMO D
_________________

I did the same, but it took me surely more than 1 minute

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

The way the question is written makes it appear more intimidating than it actually is.

Given: \(f(n) = 2^x3^y5^z\)

This means if we were given the number 729, it would be written as \(f(n) = 2^73^25^9\)

\(f(m)=9*f(v)\)
\(f(m)=3^2*f(v)\)

\(m-v=\) = \(f(m) = 2^x3^{y+2}5^z\) - \(f(v) = 2^x3^y5^z\)

This seemingly complicated function is simply telling us that the tens digit of m is 2 greater than the tens digit of v.

For example, we have two numbers 143 and 123.

143 - 123 = 20.

Answer is D.
_________________

Let ,
\(m=abc\), and

\(v=pqr\)

\(so,\)
\(f(m)= 2^a3^b5^c\)

\(f(v)= 9*2^p3^q5^r\) \(=3^22^p3^q5^r\) \(= 2^p3^{q+2}5^r\)

\(Again,\)
\(f(m)=2^a3^b5^c\)

\(So, \)
\(f(m)= 2^p3^{q+2}5^r\)

Using place value for m
\(100p+10(q+2)+r\)

\(now,\)

\(f(m)-f(v)=100p+10(q+2)+r-100p-10q-r\)

\(= 100p+10q+20+r-100p-10q-r\)

\( =20\)

Ans is 20. D
_________________

Thank you for your explanation. Just one question - how do we know that the hundreds and units digit are common for both m and v?