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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!

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    Well, just by inspection. Easy to rule out every number smaller than $11$ and $11$ clearly divides the second term but not the first. – 2017-01-21
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    What if the number had more divisors? If for some reason I couldn't/wouldn't use inspection? – 2017-01-21
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    Yes, factoring of arbitrarily complex expressions can be arbitrarily difficult. Here, though, as you have written both terms in (essentially) factored form your life is much easier. But, yes, you could write more complicated expressions which would make the task harder. – 2017-01-21
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    What are the source and motivation of this problem? – 2017-01-21

4 Answers 4

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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.

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    It is easy to check by trial and error method .You start checking from 1 and discarding and eventually reach 11.you don't even need to calculate the term.just take the number and see it divides both the parts seperately and to see that , 11 does not divide the number it ia sufficient to check that the number in the firat expression is not divisible by 11. – 2017-01-21
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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 $

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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.)

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    Your first sentence should end with $k$, not $k!$ – 2017-01-21
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    @TonyK Fixed it. Thanks. – 2017-01-21
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$2 * 3 * 4 * 5 * 6 * 7 ( 1 + 11 * 13 * 8 * 17)$

= $2 * 3 * 4 * 5 * 6 * 7 * ( 19449)$

= $2 * 3 * 4 * 5 * 6 * 7 * 3 * 3 * 2161$

2161 is not divisible by 11. So you can see 11 is the smallest number.