Find the smallest number that doesn't divide $1 * 2 * 3 * 4 * 5 * 6 * 7 + 11 * 12 * 13 * 14 * 15 * 16 * 17$ What is the smallest number that doesn't divide $1 * 2 * 3 * 4 * 5 * 6 * 7 + 11 * 12 * 13 * 14 * 15 * 16 * 17$?
I can find it programmatically brute forcing (it's $11$) but how to find it mathematically, on paper?
Thanks in advance!
 A: 2,3,4,5,6,7,8 divide one of the factors in both terms.
9 divides 3*6 in first term, 12*15 in other.
10 divides 2*5 in first, 12*15 in other.
11 divides only the second term.
Hence 11 is the answer.
A: Use prime factorization of the two summands:
$$1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7 + 11\cdot12\cdot13\cdot14\cdot15\cdot16\cdot17\\
= 2\cdot3\cdot2^2\cdot5\cdot2\cdot3\cdot7 + 11\cdot2^2\cdot3\cdot13\cdot2\cdot7\cdot3\cdot5\cdot2^4\cdot17\\
= 2^4\cdot3^2\cdot5\cdot7+2^7\cdot3^2\cdot5\cdot7\cdot11\cdot13\cdot17\\
= (2^4\cdot3^2\cdot5\cdot7)(1+2^3\cdot11\cdot13\cdot17)\\
= 2\cdot lcm(1,2,3,4,5,6,7,8,9,10)\cdot(1+2^3\cdot11\cdot13\cdot17)$$
Now we have something of the form $a\cdot lcm(1,…,p-1)\cdot(1+pb)$ where $p$ is prime (in this case, 11), and $a$'s greatest prime factor is smaller than $p$, thus $p$ does not divide $a$.
Now clearly all numbers $<p$ divide $lcm(1,…,p-1)$, but $p$ divides neither $a$, nor $lcm(1,…,p-1)$, nor $(1+pb)$. Since $p$ is prime, it does not divide the product either.
A: One fact you need here is that the product of any $k$ consecutive integers is divisible by $k$, and hence by all the primes less than $k!$.  The key features of this problem are 1. that there are two products of $k$ consecutive integers, so that every prime less than $k$ divides the sum.  2. The second product starts with the next prime $p$ larger than $k$.  So if you want a more impressive problem, pick $k = 99$ and $p=101$ to get
$$1\cdot 2 \cdots 99 + 101\cdot 102 \cdots 199.$$
Then every prime less than 101 divides both products and hence the sum.  And the next prime $101$ divides the second product, but not the first, so the answer to "What's the smallest number not dividing the sum?" is $101.$  (I suppose that sentence needs to be added that the number you're looking for is going to be prime, since if any number $n$ divides, then the prime factors of $n$ also divide and they will be smaller.)
A: $2 * 3 * 4 * 5 * 6 * 7 ( 1 + 11 * 13 * 8 * 17)$
= $2 * 3 * 4 * 5 * 6 * 7 * ( 19449)$
= $2 * 3 * 4 * 5 * 6 * 7 * 3 * 3 * 2161$
2161 is not divisible by 11. So you can see 11 is the smallest number.
