True or False?

Let us reduce the expression. $-66n^7+66n^5$ is divisible by 66. $33n^{10}-33n^3=33n^3(n^7-1)$ if n is even, $n^3$ is even and if n is odd, $n^7-1$ is even so either way $33n^{10}-33n^3$ is divisble by 66.

So we only require to check if $6n^{11}+55n^9+5n$ is divisible by 66. If n is even the expression is even and if n is odd, $55n^9+5n$ are both odd so their sum is even so the expression is divisible by 2 for all n.

$6n^{11}$ is divisible by 3. $55 \equiv 1 \pmod3, 5 \equiv 2 \pmod3$ . So $55n^9+5n \equiv n^9+2n \equiv n(n^8+2) \pmod3$ . If n is divisible by 3, $55n^9+5n$ is divisible by 3. If $n \equiv 1 \pmod3$ , $55n^9+5n \equiv n(n^8+2) \equiv 1 \times 3 \equiv 0 \pmod3$ as any number which is $1 \pmod3$ raised to any power is also $\equiv 1 \pmod 3$ .

If $n \equiv 2 \pmod3$ , $55n^9+5n \equiv n(n^8+2) \equiv 2 \times 3 \equiv 0 \pmod3$ as 8 is even and any number which is $\equiv 2 \pmod3$ raised to the power of an even number is $\equiv 1 \pmod3$ .

So, $55n^9+5n$ is always divisible by 3, so $6n^{11}+55n^9+5n$ is always divisible by 6.

So, it suffices to check whether $6n^{11}+55n^9+5n$ is divisible by 11.

$55n^9$ is divisible by 11. So it suffices to check whether $6n^{11}+5n=n(6n^{10}+5)$ is divisible by 11.

Now we have lots of case work to do.

• If $n \equiv 0 \pmod{11}$ , the expression is divisible by 11.

• If $n \equiv 1 \pmod{11}$ , $n(6n^{10}+5) \equiv 1(6+5) \equiv 0 \pmod{11}$ .

• If $n \equiv 2 \pmod{11}$ , $n(6n^{10}+5) \equiv 2(6 \times 2^{10}+5)$ . As $2^{5} \equiv -1 \pmod{11}$ , $2^{10} \equiv 1 \pmod{11}$ . So, $2(6 \times 2^{10}+5) \equiv 2(6+5) \equiv 0 \pmod{11}$ .

• If $n \equiv 3 \pmod{11}$ , $n(6n^{10}+5) \equiv 3(6 \times 3^{10}+5)$ . As $3^{5} \equiv 1 \pmod{11}$ , $3^{10} \equiv 1 \pmod{11}$ . So, $3(6 \times 3^{10}+5) \equiv 3(6+5) \equiv 0 \pmod{11}$

• If $n \equiv 4 \pmod{11}$ , $n(6n^{10}+5) \equiv 4(6 \times 4^{10}+5)$ . $4^{10} \equiv 2^{20} \equiv 1 \pmod{11}$ (from case 3). So, $4(6 \times 3^{10}+5) \equiv 4(6+5) \equiv 0 \pmod{11}$

• If $n \equiv 5 \pmod{11}$ , $n(6n^{10}+5) \equiv 5(6 \times 5^{10}+5)$ . As $5^{2} \equiv 3 \pmod{11}$ , $5^{10} \equiv 3^5 \equiv 1 \pmod{11}$ (from case 4). So, $5(6 \times 5^{10}+5) \equiv 5(6+5) \equiv 0 \pmod{11}$

• If $n \equiv 6 \pmod{11}$ , $n(6n^{10}+5) \equiv 6(6 \times 6^{10}+5)$ . $6^{10} \equiv 2^{10} \times 3^{10} \equiv 1 \pmod{11}$ (from case 3 and 4). So, $6(6 \times 6^{10}+5) \equiv 6(6+5) \equiv 0 \pmod{11}$

• If $n \equiv 7 \pmod{11}$ , $n(6n^{10}+5) \equiv 7(6 \times 7^{10}+5)$ . As $7^3 \equiv 2 \pmod{11}$ . $7^{10} \equiv 2^3 \times 7 \equiv 56 \equiv 1 \pmod{11}$ . So, $7(6 \times 6^{10}+5) \equiv 7(6+5) \equiv 0 \pmod{11}$

• If $n \equiv 8 \pmod{11}$ , $n(6n^{10}+5) \equiv 8(6 \times 8^{10}+5)$ . $8^{10} \equiv 2^{30} \equiv 1 \pmod{11}$ (from case 3). So, $8(6 \times 8^{10}+5) \equiv 8(6+5) \equiv 0 \pmod{11}$

• If $n \equiv 9 \pmod{11}$ , $n(6n^{10}+5) \equiv 9(6 \times 9^{10}+5)$ . $9^{10} \equiv 3^{20} \equiv 1 \pmod{11}$ (from case 4). So, $9(6 \times 9^{10}+5) \equiv 9(6+5) \equiv 0 \pmod{11}$

• If If $n \equiv 10 \equiv -1 \pmod{11}$ , $n(6n^{10}+5) \equiv 10(6 \times 10^{10}+5)$ $10^{10} \equiv -1^{10} \equiv 1 \pmod{10}$ . So, $10(6 \times 10^{10}+5) \equiv 10(6+5) \equiv 0 \pmod{11}$

Thus we conclude that $n(6n^{10}+5)$ is divisible by 11 for all n.

Thus, $6n^{11}+33n^{10}+55n^9-66n^7+66n^5-33n^3+5n$ is divisible by 66 for all n.

Phew!!!