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Intern
Status: Target 760
Joined: 20 Aug 2014
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Schools: Kellogg '21 (D), Kellogg 1YR '19, Booth '21 (D), Ross '21 (D), Stern '21 (S), Yale '21 (S), Darden '21 (S), Johnson '21 (S), LBS '21 (S), INSEAD Jan '19, Oxford"19, Cambridge"19, IMD '20 (S), IESE '21 (D)
GMAT 1: 730 Q49 V40
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Re: The function f is defined for each positive threedigit
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15 Oct 2017, 08:18
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 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, MV=10(x+2)10x = 20



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Re: The function f is defined for each positive threedigit
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19 Nov 2017, 00:20
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, 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 threedigit
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18 May 2018, 23:24
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, 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 threedigit
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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 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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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 'ahhah' 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




Re: The function f is defined for each positive threedigit
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