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The function f(m) is defined for all positive integers m as
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26 Jan 2011, 13:00
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The function f(m) is defined for all positive integers m as the product of m + 4, m + 5, and m + 6. If n is a positive integer, then f(n) must be divisible by which one of the following numbers? (A) 4 (B) 5 (C) 6 (D) 7 (E) 11
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Re: Defined Functions
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26 Jan 2011, 13:38
PASSINGGMAT wrote: The function f(m) is defined for all positive integers m as the product of m + 4, m + 5, and m + 6. If n is a positive integer, then f(n) must be divisible by which one of the following numbers? (A) 4 (B) 5 (C) 6 (D) 7 (E) 11
Can you explain how to answer this please. Thank you. Given: \(f(n)=(n+4)(n+5)(n+6)\), where \(n\) is a positive integer. Question: \(f(n)\) must be divisible by which one of the following numbers. Now, \((n+4)(n+5)(n+6)\) is the product of 3 consecutive integers so out of them one is definitely divisible by 3 and at least one is divisible by 2, so \(f(n)\) must be divisible by 2*3=6. Answer: C. Generally out of ANY \(k\) consecutive integers one is always divisible by \(k\) and at least one by \(k1\), \(k2\), ... For example out of ANY 5 consecutive integers there is one which is divisible by 5, and at least one which is divisible by 4, 3, and 2. That's because an integer divided by an integer \(k\) can give a remainder of: 0 (when it's divisible by \(k\)), 1, 2, ..., or \(k1\) (total of \(k\) different remainders from 0 to \(k1\)), so out of \(k\) consecutive integers there definitely will be one which gives a reminder of zero, so divisible by \(k\). Which give us the following property: the product of \(k\) consecutive integers is always divisible by \(k!\), so by \(k\) too. For example: given \(k=4\) consecutive integers \(\{3,4,5,6\}\) > the product of 3*4*5*6 is 360, which is divisible by 4!=24. If we apply this property to the original question we'll have that the product of given 3 consecutive integers \((n+4)(n+5)(n+6)\) must be divisible by 3!=6. Hope it's clear.
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Re: Defined Functions
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27 Jan 2011, 21:45
awesome explanation



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Re: Defined Functions
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29 Jan 2011, 11:59



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Re: Defined Functions
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08 Mar 2011, 19:14
Product of Any three consecutive integers must be divisible by 6.



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Re: The function f(m) is defined for all positive integers m as
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29 Mar 2015, 10:19
Bunuel Very informative explanation Bunuel! Thanks !
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Re: The function f(m) is defined for all positive integers m as
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29 Mar 2015, 18:21
Hi All, This question is based on a couple of Number Properties; as such, you can also TEST VALUES to get to the solution. We're told that the f(M) = (M+4)(M+5)(M+6) for all POSITIVE integers. We're asked which of the following numbers MUST divide into f(N). IF.... N = 1 f(1) = (5)(6)(7) At this point, you can either multiply out the numbers and check the 5 answer choices against that product OR prime factor the f(1).... (5)(6)(7) = (5)(2)(3)(7) Of the 5 answer choices, only 2 of them divide into this product (Answers B and C; 5 and 6). From here, we should look to try to eliminate one of the options. It's actually not that hard.... IF.... N = 2 f(2) = (6)(7)(8) Looking at this, we can see that 6 IS a factor while 5 is NOT. Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: The function f(m) is defined for all positive integers m as
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30 Sep 2016, 21:33
[align=]I tried the sum with f(13), which is equal to 17 X 18 X 19 (this number isn't divisible by 6)?[/align]
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Re: The function f(m) is defined for all positive integers m as
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01 Oct 2016, 15:45
Hi pnh0505, You are correct that when you use M=13, then f(13) = (17)(18)(19). However, that product IS divisible by 6  because 18 is divisible by 6. GMAT assassins aren't born, they're made, Rich
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Re: The function f(m) is defined for all positive integers m as
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01 Mar 2017, 00:24
In three consecutive integers we can say that one of it will be an even number and one will a multiple of three. To the product will be definitely the product of 6. Option C
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