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Re: The function f is defined for each positive three-digit
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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, 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 three-digit
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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.
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 three-digit
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07 Dec 2017, 14:17
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
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Re: The function f is defined for each positive three-digit
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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
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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.
Answer D Posted from my mobile device
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Re: The function f is defined for each positive three-digit
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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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Big Fat Disclaimer: I am NotAnExpert, but rather, merely a person that is studying for the GMAT and wants to make the best of, and take the most out of this forum. My only goal is to contribute to the community from the which I have taken so much, and ultimately help make the GMAT Club forums a better place to learn. In my posts, I share my learning outcomes by verbalizing how I solved problems, and hope that the added perspective can help anyone achieve an 'ah-hah' moment.
Sometimes it takes a concept being explained in 15 different ways to achieve an 'ah hah' breakthrough moment and I am here to contribute one, of hopefully many, unique perspectives. I do not encourage or participate in posts that simply state "that was easy", or "relevant, out of scope, correct". I find that people - myself included - often have a difficult time truly understanding the fact that in CR/RC questions, there will be one very definitively black and white correct answer, just a there is in Quant. As a result, my posts are exclusively focused on CR and RC. There is no such thing as a 'kinda right' question and because of this, I contribute detailed posts on how I came to my answer in hopes that it will connect with someone.
Wishing everyone all the best
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Re: The function f is defined for each positive three-digit
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26 Jul 2019, 14:26
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