Let $n$ be a positive integer relatively prime with $10$. Prove that the digits of hundreds of $n^{20}$ is even.

Question

Let $n$ be a positive integer relatively prime with $10$. Prove that the hundreds digit of $n^{20}$ is even.

I know this has something to do with $\bmod 1000$, I'm just not sure how to write a proof on it. Thank you!

Answer 1

Note that $x^{20} \equiv_{25} 1$ and $x^{20} \equiv_8 1$. So $x^{20} \equiv_{25×8} 1$. Can you finish from here.

Answer 2

Here is an alternative which does not use totient function as in Mike's answer, instead it relies heavily on the binomial expansion (used three times in this proof).

Notice that for $a\in\{0,1,\cdots,9\}$ and $b>0$

$x^{20}=(a+10b)^{20}=\sum\limits_{k=0}^{20} \binom{20}{k}a^{20-k}10^k=a^{20}+200(\cdots)$

This is because $200\mid 10^k\binom{20}{k}$ for $k\ge 1$

• for $k=1$ this is $10\times 20=200$
• for $k\ge 2$ then $100\mid 10^k$ and $2\mid \binom{20}{k}$

As a consequence we only have to examine values $a=1,3,7,9$ because if a number ends with $a=0,2,4,5,6,8$ it is either divisible by $2$ or $5$ which is not possible since $\gcd(x,10)=1$.

• $1^{20}\equiv 1\pmod{200}$
• $3^{20}=81^{5}\equiv (80+1)^5\equiv 1^5+200(\cdots)\equiv 1\pmod{200}$
• $9^{20}=3^{20}\,3^{20}\equiv 1\pmod{200}$
• $7^{20}=2401^{5}\equiv (2400+1)^5\equiv 1^5+200(\cdots)\equiv 1\pmod{200}$

This means $x^{20}=200y+001$ so the hundreds digit is even.