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If n is a positive integer, is (n-1)(n)(n+1) a multiple of 24?

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If n is a positive integer, is (n-1)(n)(n+1) a multiple of 24?  [#permalink]

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New post 13 Apr 2017, 11:38
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If n is a positive integer, is (n-1)(n)(n+1) a multiple of 24?

1) (n-1) is an even integer
2) n is a multiple of 3

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If n is a positive integer, is (n-1)(n)(n+1) a multiple of 24?  [#permalink]

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New post 13 Apr 2017, 12:21
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If n is a positive integer, is (n-1)(n)(n+1) a multiple of 24?

Those are 3 consecutive positive integers.

1) (n-1) is an even integer

As per statement we have the situation: Even, ODD, Even

Any three consecutive integers, mean we have guarantee that we at least 3.

Two even integers, we can guarantee that we have 8.

So answer to the question ins always yes.

Another way is plug numbers.

n-1 =2 & n=3 & n+1=4............2*3*4=24

n-1 =4 & n=5 & n+1=46............4*5*6=120

n-1 =6 & n=7 & n+1=8............6*7*8=336

You can notice above rule mentioned in first solution.

Sufficient

2) n is a multiple of 3

Plug in numbers.

n-1 =2 & n=3 & n+1=4............2*3*4=24............Answer is Yes

n-1 =5 & n=6 & n+1=7............5*6*7=210..........Answer is No

insufficient

Answer: A
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Re: If n is a positive integer, is (n-1)(n)(n+1) a multiple of 24?  [#permalink]

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New post 13 Apr 2017, 12:33
Answer is A.
Question stem says (n-1), n and (n+1) are consecutive positive integers. Hence, one of them is a multiple of 3

Stat1: alone is sufficient
(n-1) is even implies that (n-1) is a multiple of 2 or 4
if (n-1) is multiple of 2, then (n+1) is a multiple of 4
if (n-1) is multiple of 4, then (n+1) is a multiple of 2

There is a multiple of 2 and a multiple of 4 in the set {(n-1) ; (n+1)}

as per question stem, There is a multiple of 3 in the set {(n-1) ; n ; (n+1)}
Therefore, (n-1)*n *(n+1) is multiple of 2*4*3 = 24

Stat2 is not sufficient
n=3 gives YES answer
n=6 gives NO answer
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If n is a positive integer, is (n-1)(n)(n+1) a multiple of 24?  [#permalink]

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New post 06 May 2017, 02:02
Also consider the following case:

n-1 = 0, n=1, n+1 = 2; (coz, "0" is an "even integer")

i.e (n-1)(n)(n+1) = 0, which is divisible by any number, hence a multiple of any number.

The answer is "A" .
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Re: If n is a positive integer, is (n-1)(n)(n+1) a multiple of 24?  [#permalink]

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New post 17 Aug 2018, 18:20
since n-1 is even, n+1 has to even too. the product of n-1 and n+1 should be multiple of 8 [2m*(2m+2))=4m(m+1),m is positive int]
as we know the product of three consecutive positive int should be multiple of 3, so in this case it should be multiple of 24
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Re: If n is a positive integer, is (n-1)(n)(n+1) a multiple of 24? &nbs [#permalink] 17 Aug 2018, 18:20
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If n is a positive integer, is (n-1)(n)(n+1) a multiple of 24?

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