Q13/029) Prove that 21n^22+22n^26+34n^32≡0(mod 77)

we have 77 = 7 * 11
as 7 annd 11 are co-primes
so a mod 77= 0 if a mod 7 = 0 and a mod 11 = 0

21 mod 7 = 0 , 22 mod 7 = 1, 34 mod 7 = -1

so 21n^22+22n^26+34n^32 = n^ 26 - n^ 32 = n^26(1 - n^6)

if n = 0 mod 7 the n^26 = 0 mod 7 else n is co-prime to 7

so n^6 =1 mod 7 ( by Fremat's Little theorem)
so
21n^22+22n^26+34n^32≡0 mod 7 ..1

now for mod 11
21 = -1 mod 11 , 34 = 1 mod 11, and 22 = 0 mod 11

so
21n^22+22n^26+34n^32
= - n^22 + n^ 32 = n^22 (n^10-1)

if n = 0 mod 11 then n^22 = 0 mod 11

or n is corpime to 11 so n^ 10 -1 = 0 mod 11

so 21n^22+22n^26+34n^32 = 0 mod 11...2

so 21n^22+22n^26+34n^32≡0(mod 77)  (from 1 and 2)