divisibility of N from 1,2,3,...k Good day! I encountered this problem yesterday and I eventually answered the problem correctly.
The problem is this:If N is divisible by 1,2,3,. . . 13, then N must also be divisible by 14 and 15.
Using this same idea, what is the smallest integer  such that the following statement is true?
If N is divisible by 1,2,3,. . . M, then N must also be divisible by M + 1, M + 2, M + 3 and M + 4.
But I am wondering if there is a general rule/ proof/ algorithm for this? I used rough way to solve this one,, 
 A: Your statement is true if and only if each of $M+1,M+2,M+3,M+4$ is not a prime power.
I doubt there is a good way to find $M$ other than brute force. $M=53$ give $54,55,56,57$ without any prime powers, so that $M$ works, but I haven't found if it is the smallest.
More generally, for $M+1,M+2,\dots,M+k$, you'd need to know something about prime power gaps. I'm not sure what is known about that, but I'm assuming they are arbitrarily large.
For general $k$, I think you can use $M=(k+1)!^2+1$. Then $M+1,M+2,\dots,M+k$ are not prime powers, but this $M$ is not the smallest such $M$ by far.
A smaller $M$ for a general $k$ is:
$$M=1+\prod_{p\leq k+1} p^{1+\lfloor \log_p (k+1)\rfloor}$$
This gives $M=13$ for $k=2$ and $M=1+8\cdot 9=73$ for $k=3$.
A: hint
If one of the four numbers $M+1, M+2, M+3 , M+4$ is prime, the statement is not true.
the primes are
$2,3,5,7,11,13,17,19,23,29,31,37... $
So, the possible values of $M $ are
$$23,31,32,47,53$$
you check them one by one until 53 which works.
A: $144\,403\,552\,893\,600=2^5\times 3^3\times 5^2\times 7\times 11\times 13\times 17\times 19\times 23\times 29\times 31$
Is divisible by $2,\;3,\;\ldots,31$ and by $32,\;33,\;35,\;36$ and is the smallest number with such a property
It is a product of primorials, BTW...
