Power of 11!

Consider modulo $23$ . If any of the number is divisible by $23$ , then the expression is automatically divisible by $23$ . Suppose that none of them is divisible by $23$ ,Now we want to show that it is divisible by $23$ .Note that the any number raised to power $11$ must leave a remainder of $1$ or $10$ when divided by $23$ .

Its proof: By Fermat's little theorem , $a^{22} \equiv 1\pmod{23} \\ \implies a^{11} \equiv 1~ \text{or}~ -1 \pmod{23}$ This shows that any number when raised to power of $11$ will leave a remainder $1~ \text{or}~ 10$ when divided by $23$ . Now since the numbers are three and possible remainders are only two,there will be two numbers which leave same remainder, suppose that $WLOG$ , $a^{11}$ and $b^{11}$ leave the same remainder when divided by $23$ , This implies that $a^{11}-b^{11} \equiv 0 \pmod{23}$ , and hence the given expression is always divisible by $23$ .

If we consider the cases: $\begin{array}{llll} a & b & c & N(a,b,c) \\ 1 & 2 & 3 & 2^2 \times 3 \times 23^4 \times 89 \times 331 \times 3851 \\ 1 & 2 & 4 & 2^{14} \times 3 \times 23^3 \times 89^3 \times 683 \\ 1 & 9 & 19 & 2^5 \times 3^4 \times 5 \times 19 \times 23 \times 67 \times 661 \times 3851 \times 104281 \times 62060021 \times 11645887783861 \end{array}$ we see that the largest number that can divide all these numbers $N(a,b,c)$ is $2^2 \times 3 \times 23$ . It remains to show this is the largest.

• If two or three of the numbers $a,b,c$ are even, then $abc$ is a multiple of $4$ , and so $N(a,b,c)$ is a multiple of $4$ . If all of the numbers $a,b,c$ are odd, then all the numbers $a^{11}$ , $b^{11}$ , $c^{11}$ are odd, and so all their differences will be even, and hence $N(a,b,c)$ will be a multiple of $8$ . If one of the numbers $a,b,c$ is even, then one of the differences $a^{11} - b^{11}$ , $b^{11} - c^{11}$ , $c^{11} - a^{11}$ will be even, and so $N(a,b,c)$ will be a multiple of $4$ . Thus $N(a,b,c)$ is always a multiple of $4$ .
• If any of the numbers $a,b,c$ are multiples of $3$ , then $N(a,b,c)$ is a multiple of $3$ . If not then $a^2 \equiv b^2 \equiv c^2 \equiv 1 \pmod{3}$ , so that $a^{11} \equiv a, b^{11} \equiv b, c^{11} \equiv c \pmod{3}$ . Each of $a^{11},b^{11},c^{11}$ is congruent to $1$ or $2$ modulo $3$ and so at least one of the differences $a^{11} - b^{11}$ , $b^{11} - c^{11}$ , $c^{11} - a^{11}$ will be a multiple of $3$ , and so $N(a,b,c)$ will be a multiple of $3$ . Thus $N(a,b,c)$ is always a multiple of $3$ .
• If any of the numbers $a,b,c$ are multiples of $23$ , then $N(a,b,c)$ is a multiple of $23$ . If not then $a^{22} \equiv b^{22} \equiv c^{22} \equiv 1 \pmod{23}$ , and so each of $a^{11},b^{11},c^{11}$ is congruent to either $1$ or $-1 \equiv 22$ modulo $23$ . Thus at least one of the differences $a^{11} - b^{11}$ , $b^{11} - c^{11}$ , $c^{11} - a^{11}$ will be a multiple of $23$ , and so $N(a,b,c)$ will be a multiple of $23$ . Thus $N(a,b,c)$ is always a multiple of $23$ .

Thus the largest integer that divides all the numbers $N(a,b,c)$ is $2^2 \times 3 \times 23 = \boxed{276}$ .