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Factorials, divisibility & composite doubt

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Intern
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Joined: 18 Mar 2019
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Factorials, divisibility & composite doubt  [#permalink]

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New post 17 Jul 2019, 00:18
Hi. I have a bit of a vague question if anybody can help fill in the blanks.

1) Is 5!+2 a composite? Why?
2) How is 5!+3 divisible by 3?
3) is 6!+3 a composite?

I do understand it when I open up 5! But if someone could help me with the logic of figuring this out, I'd be very grateful.

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Director
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Joined: 20 Jul 2017
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Location: India
Concentration: Entrepreneurship, Marketing
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Re: Factorials, divisibility & composite doubt  [#permalink]

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New post 17 Jul 2019, 05:44
2
Dhwani12 wrote:
Hi. I have a bit of a vague question if anybody can help fill in the blanks.

1) Is 5!+2 a composite? Why?
2) How is 5!+3 divisible by 3?
3) is 6!+3 a composite?

I do understand it when I open up 5! But if someone could help me with the logic of figuring this out, I'd be very grateful.

Posted from my mobile device


5! = 1x2x3x4x5
So, if you agree that 5! Can be written as 2k, for some positive integer k
Or
3m, for some positive integer m
And
6! Can be written as 3n, for some positive integer n

1) Is 5!+2 a composite? Why?
—> 2k + 2 = 2(k + 1)
—> factors are 1, 2, k+1 & 2(k+1)
So, composite

2) How is 5!+3 divisible by 3?
—> 3m + 3 = 3(m+1)
—> factors are 1, 3, m+1 & 3(m+1)
So, composite

3) is 6!+3 a composite?
—> 3n + 3 = 3(n+1)
—> factors are 1, 3, n+1 & 3(n+1)
So, composite
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Factorials, divisibility & composite doubt  [#permalink]

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New post 10 Aug 2019, 06:33
5! is a multiple of 2, 3, 4, and 5. This means 5! + 3 is a multiple of 3 plus 3, therefore it is a multiple of 3.
5!+2 is even because 5! and 2 are even, and even + even = even. This means 5!+2 is divisible by at least 3 factors: 5!+2, 1, 2. Thus it is composite.
6!+3 is a multiple of 3 + 3, so it is divisible by at least 3 factors: 6!+3, 3, and 1. Thus it is also composite.
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Factorials, divisibility & composite doubt   [#permalink] 10 Aug 2019, 06:33
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