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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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18 Nov 2017, 23:20
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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21 Nov 2017, 21:17
saxenashobhit wrote:
f(m)=9f(v) can be written as
f(m) = $$3^2$$f(v).

So tenth place of m is 2 number greater than tenth place of v. This means m is 20 greater than v.

love it when someone posts such effective and minimalistic explanations! Kudos to you
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Re: The function f is defined for each positive three-digit  [#permalink]

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07 Dec 2017, 13:17
1
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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18 May 2018, 22:24
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  [#permalink]

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16 May 2019, 14:38
Hi there

I’m not sure I understand the underlying logic. Could you explain more please? Thank you!
saxenashobhit wrote:
f(m)=9f(v) can be written as
f(m) = $$3^2$$f(v).

So tenth place of m is 2 number greater than tenth place of v. This means m is 20 greater than v.

Posted from my mobile device
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Re: The function f is defined for each positive three-digit  [#permalink]

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26 Jul 2019, 13:26
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)
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Re: The function f is defined for each positive three-digit  [#permalink]

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10 Apr 2020, 06:16
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

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
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Re: The function f is defined for each positive three-digit   [#permalink] 10 Apr 2020, 06:16

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