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Re: The function f is defined for each positive threedigit
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18 Nov 2017, 23:20
VeritasPrepKarishma wrote: Orange08 wrote: The function f is defined for each positive threedigit 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 threedigit positive integers such that f(m)=9f(v), them mv=? (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 threedigit integer n by \(f(n) = 2^x3^y5^z\) , where x, y and z are the hundreds, tens, and units digits of n, respectivelyI 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 threedigit
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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.
Answer D 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 threedigit
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07 Dec 2017, 13:17
Hi All, While this prompt is 'complexlooking', 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 3digit 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 3digit numbers such that f(M) = 9(f(V)). We're asked for the value of MV. Given how the function multiplies 'powers of primes', for one 3digit 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
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Re: The function f is defined for each positive threedigit
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18 May 2018, 22:24
Orange08 wrote: The function f is defined for each positive threedigit 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 threedigit positive integers such that \(f(m)=9*f(v)\) , then \(mv=\) ?
(A) 8 (B) 9 (C) 18 (D) 20 (E) 80 This Q can be easily solved within 1 minute by the substitutionmethod. As soon as i read "positive threedigit 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 mv = 120100 = 20
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Re: The function f is defined for each positive threedigit
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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.
Answer D Posted from my mobile device



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Re: The function f is defined for each positive threedigit
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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 3digit number of 202 To have f(n)=f(v)=900, I set x=2, y=2, and z=2 resulting in a 3digit 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 threedigit
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10 Apr 2020, 06:16
Orange08 wrote: The function f is defined for each positive threedigit 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 threedigit positive integers such that \(f(m)=9*f(v)\) , then \(mv=\) ?
(A) 8 (B) 9 (C) 18 (D) 20 (E) 80 The function f is defined for each positive threedigit 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 threedigit positive integers such that \(f(m)=9*f(v)\) , then \(mv=\) ? 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 threedigit
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