$S= 2^{1} + 22^{11} + 222^{111} \cdots +22222222222^{11111111111}$. Remainder if $S$ is divided by $7?$ Prove that $S$ is not divisible by $5.$

Question

Let, $S= 2^{1} + 22^{11} + 222^{111} \cdots +22222222222^{11111111111}$.

1. What will be the remainder if $S$ is divided by $7$?

2. Prove that $S$ is not divisible by $5.$

Attempt:

• Can the first part be solved by using the Chinese Remainder Theorem and modular arithmetic?
• Does anyone have any other clue or hint to solve the problem?

Answer 1

For the last part: taking everything modulo $5$ and noting that bases are $\equiv 2$ modulo $5$ while we only have to consider the exponents modulo $4$ because of Fermat's little theorem, so we get remainders mod 5 of $2^1 = 2$, and $2^{3} \equiv 3$ for the other ones (as all exponents are 11 mod 100, so 3 mod 4), so $10 \times 3$ for the last terms and so 32 in total which is still $2$ modulo $5$ and not $0$ so the sum is not divisible by $5$.

For the first part look at modulo $7$ values. The bases are (mod 7):

$$2,1,5,3,4,0,2,1,5,3,4$$

while the exponents (mod 6, because of little Fermat) are:

$$1,5,3,1,5,3,1,5,3,1,5$$

so consider $$2^1+ 1^5 + 5^3 + 3^1+4^5+ 0^1+2^5+1^3+5^1+4^5 \pmod{7}$$

which equals $0$. So the number is divisible by $7$.