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# The function f is defined for each positive three-digit

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Intern
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Joined: 20 Aug 2014
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Re: The function f is defined for each positive three-digit  [#permalink]

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15 Oct 2017, 08:18
1
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

2^x3^y5^z is the function: If you caught the above highlighted clue from the question stem you are through. Since the differential is of 3^2 means that at the tens position, 10*2 is the additional figure to the overall number, m. Thus, M-V=10(x+2)-10x = 20
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Re: The function f is defined for each positive three-digit  [#permalink]

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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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21 Nov 2017, 22:17
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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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  [#permalink]

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16 May 2019, 15:38
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, 14:26
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] 26 Jul 2019, 14:26

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