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The function f(m) is defined for all positive integers m as

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The function f(m) is defined for all positive integers m as  [#permalink]

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Updated on: 15 Jul 2019, 03:01
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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

Source: Nova GMAT
Difficulty Level: 600

Originally posted by PASSINGGMAT on 26 Jan 2011, 14:00.
Last edited by SajjadAhmad on 15 Jul 2019, 03:01, edited 1 time in total.
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Re: The function f(m) is defined for all positive integers m as  [#permalink]

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26 Jan 2011, 14:38
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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

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.

Generally out of ANY $$k$$ consecutive integers one is always divisible by $$k$$ and at least one by $$k-1$$, $$k-2$$, ... 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 $$k-1$$ (total of $$k$$ different remainders from 0 to $$k-1$$), 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: The function f(m) is defined for all positive integers m as  [#permalink]

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27 Jan 2011, 22:45
awesome explanation
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Re: The function f(m) is defined for all positive integers m as  [#permalink]

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29 Jan 2011, 12:59
Ye, Bunuel - u rocked it. +1.
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Re: The function f(m) is defined for all positive integers m as  [#permalink]

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08 Mar 2011, 20:14
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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  [#permalink]

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29 Mar 2015, 11:19
Bunuel

Very informative explanation Bunuel!

Thanks !
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Re: The function f(m) is defined for all positive integers m as  [#permalink]

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29 Mar 2015, 19:21
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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.

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Re: The function f(m) is defined for all positive integers m as  [#permalink]

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30 Sep 2016, 22:33
[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  [#permalink]

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01 Oct 2016, 16:45
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.

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Re: The function f(m) is defined for all positive integers m as  [#permalink]

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01 Mar 2017, 01:24
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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Re: The function f(m) is defined for all positive integers m as  [#permalink]

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03 Apr 2019, 23:00
Bunuel wrote:
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

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.

Generally out of ANY $$k$$ consecutive integers one is always divisible by $$k$$ and at least one by $$k-1$$, $$k-2$$, ... 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 $$k-1$$ (total of $$k$$ different remainders from 0 to $$k-1$$), 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.

Best Explanation !!
Re: The function f(m) is defined for all positive integers m as   [#permalink] 03 Apr 2019, 23:00
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