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If n is a positive integer less than 100 and n^2 + 5n + 6 is a multipl

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Bunuel
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If n is a positive integer less than 100 and n^2 + 5n + 6 is a multipl [#permalink] New post   14 Apr 2020, 09:31
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If n is a positive integer less than 100 and \(n^2 + 5n + 6\) is a multiple of 6, then how many values can n take?

A. 33
B. 65
C. 66
D. 67
E. 68


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C
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If n is a positive integer less than 100 and n^2 + 5n + 6 is a multipl [#permalink] New post   Updated on: 15 Apr 2020, 00:41
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Bunuel wrote:
If n is a positive integer less than 100 and \(n^2 + 5n + 6\) is a multiple of 6, then how many values can n take?

A. 33
B. 65
C. 66
D. 67
E. 68


Are You Up For the Challenge: 700 Level Questions


n^2 + 5n + 6 = Multiple of 6

i.e. n^2 + 5n also should be multiple of 6

i.e. n(n+5) must be a multiple of 6

So n can be 1, 4, 7, 10, 13 ... (values at a gap of 3) and so on for (n+5) to be divisible by 6 i.e. 33 values

OR

So n can be 3, 6, 9, 12, 15, 18 ... (values at a gap of 3) and so on for (n) to be divisible by 6 i.e. 33 values


Total Values = 33+33 = 66

Answer: Option C

Originally posted by GMATinsight on 14 Apr 2020, 09:40.
Last edited by GMATinsight on 15 Apr 2020, 00:41, edited 1 time in total.
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Re: If n is a positive integer less than 100 and n^2 + 5n + 6 is a multipl [#permalink] New post   14 Apr 2020, 10:29
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n2 + 5n + 6
=(n + 2) (n + 3)
(n + 2) and (n + 3) are two consecutive numbers.
One of (n + 2) or (n + 3) has to be even.
We need n such that (n + 2) (n + 3) is a multiple of 3.
(n + 2) or (n + 3) should be a multiple of 3.
n leaves a remainder of 0 or 1 when divided by 3.
n could be 1, 3, 4, 6, 7, 9, 10, 12, 13, 15, 16, 18 ....
So, from among the first 99 natural numbers, N cannot take 2, 5, 8, 11, 14…….98.
There are 33 numbers in this list
we are effectively listing all numbers {3 × 0 + 2, 3 × 1 + 2, 3 × 3 + 2, ….3 × 32 + 2}.
So, N can take 99 – 33 = 66 values.
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Re: If n is a positive integer less than 100 and n^2 + 5n + 6 is a multipl [#permalink] New post   14 Apr 2020, 11:25
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Bunuel wrote:
If n is a positive integer less than 100 and \(n^2 + 5n + 6\) is a multiple of 6, then how many values can n take?

A. 33
B. 65
C. 66
D. 67
E. 68


Are You Up For the Challenge: 700 Level Questions


Asked: If n is a positive integer less than 100 and \(n^2 + 5n + 6\) is a multiple of 6, then how many values can n take?

n^2 + 5n + 6 = (n+2)(n+3)
Since (n+2)(n+3) is a multiple of 2
Either n+2 = 3k or n+3 = 3k
Either n = 3k or 3k + 1 form

n = {1,3,4,6,......96,97,99}
Number of terms of 3k+1 series = (97-1)/3+1 = 32 +1 = 33
Number of terms of 3k series = (99-3)/3+1 = 32 +1 = 33

Number of values n may take = 33 + 33 = 66

IMO C
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If n is a positive integer less than 100 and n^2 + 5n + 6 is a multipl [#permalink] New post   15 Apr 2020, 09:01
Bunuel wrote:
If n is a positive integer less than 100 and \(n^2 + 5n + 6\) is a multiple of 6, then how many values can n take?

A. 33
B. 65
C. 66
D. 67
E. 68


Are You Up For the Challenge: 700 Level Questions


\(n^2 + 5n + 6\) = 6*a
can be written as
6*a= n*(n+5)+6
note n<100
we see that at n odd integer value its true so total odd integers 50
and also at n even values i.e factor of 6 i.e 6,12,18,24 which are total 16
so total 50+16 ; 66
OPTION C
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If n is a positive integer less than 100 and n^2 + 5n + 6 is a multipl [#permalink]
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