Derek's multi-divisible integer

Let $x \in \{N,N+1,N+2\} | 12 \mid x$ . Then $x \cong 0 \pmod{12} \to x \cong 0 \pmod {3}$ . This means that $x \pmod{9}$ is one of $0,3,6$ , and so it is clearly $0$ , because if $x \not\cong 0 \pmod{9}$ , none of $N,N+1,N+2$ will be divisible by $9$ . Similarly, we get $x \cong 0 \pmod{8}$ and so $x \cong 0 \pmod{72}$ .

$\\$

We consider the multiples of $72$ modulo $11$ . Beginning with $72 \cdot 1$ , the remainders are $6,1,7,2,8,3,9,4,10,5,0,6,...$ . Thus, if $x = 72a$ , $a$ must be one of $0,2,4,7,9$ modulo $11$ , or else $11$ can not possibly divide any of $N,N+1,N+2$ .

Finally, we note that $x$ must be one of $0,1,2,8,9$ modulo $10$ , and so $a$ must be one of $0,1,4,5,6,9$ modulo $10$ , or else $10$ can't divide any of $N,N+1,N+2$ . We proceed to check values of $a$ from above, and we see that $x = 72\cdot11 = 792$ satisfies $x \cong 0 \pmod{72},$ $x \cong 0 \pmod{11}, x \cong 2 \pmod{10}$ and also satisfies $x \cong 1 \pmod{7}$ , and so $N = \fbox{790}$ is the smallest possible value of $N$ which satisfies the given conditions.

P.S. In case the last few lines were not clear, we don't need to check every value up to $12$ for divisibility. Checking divisibility by $72,11,10,7$ is sufficient to ensure than $N$ satisfies the given conditions. Also, since $x \cong 2 \pmod{10}$ , it must be equal to $N+2$ .

The two highest numbers of M are 11 and 12. Therefore we must find the LCM of them because this is the lowest possible option of N (132). Because there is N, N+1 and N+2 we must make sure as we go through the common multiples of 11 and 12 that they are with 2 numbers of a multiple of 10. So we try 130. It's not a multiple of 3 so it can't work. Next we go to 264, then 396. These aren't within 2 of a 10 so we leave them. Next is 528 which is 2 off 530. However, 530 isn't a multiple of 3 so it can't work. Next is 660 which isn't a multiple of 7 so it can't work. Next is 790 which is a multiple of everything so that's the answer.

Let n = N + 1. Then n must satisfy n^3 - n = lcm(1,2,...,12) * k, for some constant k (note that this is not a sufficient condition, it is a necessary condition). Thus n^3 - n = 27720k. From here it is purely guess and check. The least integer solution for n would be calculated to be 791, hence N = 790.