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If $n = 51! +1$, Then find no of primes among $n+1,n+2,\ldots, n+50$

Really speaking, I don't have any clue ...

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    $\begingroup$ Can you prove that every one of the numbers has a factor? $\endgroup$ – Daniel Fischer Jun 30 '14 at 20:07
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    $\begingroup$ But I think there is no prime as: n+1 = 51! +1 + 1 = 51!+2 = 2 ( (51!/2) + 1 ) As 51!/2 is integer, hence 2 is a factor Likewise each term has integral factors $\endgroup$ – Arnab Dutta Jun 30 '14 at 20:08
  • $\begingroup$ yes that's right $\endgroup$ – John Fernley Jun 30 '14 at 20:09
  • $\begingroup$ What factor has $n+2$? Which $n+3$? And so on. Do it systematically, and you've solved it in no time. $\endgroup$ – Daniel Fischer Jun 30 '14 at 20:10
  • $\begingroup$ thanks...it was a starting problem otherwise it is very easy...my bad :) $\endgroup$ – Arnab Dutta Jun 30 '14 at 20:15
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First you need to review the definition of the factorial: $$m! = \prod_{i = 1}^m i = 1 \times 2 \times 3 \times \cdots \times m.$$ This means that $m!$ is divisible by 2, by 3, by 4, by every number up to $m$.

Therefore $51!$ is divisible by 2, by 3, by 4 and by every number up to 51 (and a few others greater than 51, but you don't need to worry about those for this problem).

Then $51! + 2$ is also divisible by 2.

$51! + 3$ is also divisible by 3.

$51! + 4$ is also divisible by 4.

And so on and so forth to $51! + 51$, which is divisible by 51.

Maybe $51! + 1$ is prime. Maybe so is $51! + 53$. But in between those two numbers, there are zero primes.

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  • $\begingroup$ 51!+1 is not a prime. It is divisible by 1098482930441153. $\endgroup$ – VividD Jul 1 '14 at 10:29
  • $\begingroup$ @VividD She did say maybe $51! + 1$ is prime. It doesn't change the fact that there are no primes between that number and $51! + 51$. By any chance have you been able to factor $51! + 53$? $\endgroup$ – user153918 Jul 1 '14 at 13:25
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    $\begingroup$ 51! + 53 = 13337 * 665295941 * 20466432548293 * 246330851030970479 * 34674568702499841811147 @Alonso del Arte $\endgroup$ – VividD Jul 4 '14 at 1:40
  • $\begingroup$ @VividD how do you deduce this btw ? $\endgroup$ – Arnab Dutta Jul 23 '16 at 18:56
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The number $51!$ has as non-trivial factors every natural number preceding $51$.

Thus any $51! + 2$, $51! + 3$, etc. will be divisible by $2,3,4,\cdots$ respectively. $$n=51! + 1$$ $$n+a = 51! + (1+a) = \left(\frac{51!}{1+a}+1\right)\cdot (1+a)$$ If $1\leq a \leq 50$ then $\frac{51!}{1+a}$ is an integer and so $a+1$ divides $n+a$. (and they are obviously not equal, and $a+1\neq 1$).

Thus there are no primes in the range specified!

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In addition to Michael Hardy's answer, which precludes all of n+1 to n+50, you can also prove that n is in fact composite as well using Wilson's Theorem which states that a number n is prime if and only if (n - 1)! ≡ (n-1) mod n.

Here, n is (51! + 1), so (n-1)! = 51!!. Therefore, (51! + 1) is prime iff 51!! ≡ 51! mod (51! + 1). However, we now that (51!+1) is of course a factor of 51!! since (51! + 1) < 51!!, so it is impossible that 51!! would be congruent to anything other than 0 modulo (51! + 1). Therefore, 51! + 1 is also definitely not prime.

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The offset by $1$ on $51!$ may make it a little confusing. For any integer $2 \leq n \leq 51$, you will have that $51! + n$ is divisible by $n$ because $51!$ is divisible by $n$. This is an equivalent statement to your problem. Now you just need to show $51!$ is divisible by $n$ and that $n$ does not equal $51! + n$.

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