Let, $S= 2^{1} + 22^{11} + 222^{111} \cdots +22222222222^{11111111111}$.
What will be the remainder if $S$ is divided by $7$ ?
Prove that $S$ is not divisible by $5$
Attemp:
*The first part can be solved by using the Chinese Remainder Theorem and modular arithmetic.
Anyone has anyother clue or hint to solve the problem ?
Answer
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 $7$.
No comments:
Post a Comment