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The function f is defined for each positive three-digit

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Re: The function f is defined for each positive three-digit  [#permalink]

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New post 19 Nov 2017, 00:20
VeritasPrepKarishma wrote:
Orange08 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) 80


The method I would use to solve this problem is exactly the same as saxenashobhit did. I am writing it down just to give a little more detailed explanation.

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

I read the first line and say, "Well, this only means that \(f(146) = 2^1*3^4*5^6\), no matter how sordid it looks."

Now next line tells me that \(f(m)=9f(v) = 3^2 f(v)\). This tells me that m and v are the same except that the ten's digit of m is 2 more than ten's digit of v. This means m - v = 20.
If this is unclear, look at the example above. I can say that if f(v) = \(f(146) = 2^1*3^4*5^6\) then f(m) will be \(3^2 * 2^1*3^4*5^6\) i.e. \(f(m) = 2^1*3^6*5^6\) giving us m as 166. m - v = 166 - 146 = 20
Then m - v = 20

I found that Karishma's explanation above as simplest and best one, great job!
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Re: The function f is defined for each positive three-digit  [#permalink]

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New post 07 Dec 2017, 14:17
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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.

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Re: The function f is defined for each positive three-digit  [#permalink]

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New post 18 May 2018, 23:24
Orange08 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 \(f(m)=9*f(v)\) , then \(m-v=\) ?

(A) 8
(B) 9
(C) 18
(D) 20
(E) 80


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
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Re: The function f is defined for each positive three-digit &nbs [#permalink] 18 May 2018, 23:24

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