Divisibility Problem from an Olympiad Book

Question

In a practice problem set of the Olumpiad book that I am solving, the following question has me stumped. $$\text{Prove that }3(7^{200}+7^{202}+7^{204})+7(3^{200}+3^{204}) -210 \text{ is divisible by 10.}$$ My attempt is as follows

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Using the Sum of GP formula $$3\cdot7^{200}\Big(\frac{49^3-1}{49-1}\Big)+7\cdot3^{200}(\frac{81^2-1}{81-1}) -210 $$ Simplifying we get, $$ 7353\cdot7^{200}+574\cdot3^{200}-210$$

Now the proof is equivalent to showing that $$7353\cdot7^{200}+574\cdot3^{200} \equiv 0 \mod 10$$ Since both 7 and 3 are coprime to 10, by Euler's Theorem $3^4 \equiv 7^4 \equiv 1 \mod 10$
So, we get $$ 7353 + 574 \equiv 7 \mod 10 $$ This is the problem

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Any insight or an alternate approach would be helpful.

Answer 1

Well, it's false -- at least as stated.

Easy way to check that is look at parity (mod 2). $7^{200} + 7^{202} + 7^{204}$ is odd + odd + odd, so is odd. $3^{200} + 3^{204}$ is odd + odd, so is even. 3 * odd is odd, 7 * even is even. odd + even is odd, odd - 210 is odd. So your overall result is odd, and thus not divisible by 10.

Answer 2

Much easier, noting that $7^2 = 49 \equiv -1 \pmod{10}$ and $3^2 = 9 \equiv -1 \pmod{10}$, and considering the effect of raising $-1$ to odd and even powers, you can almost immediately reduce the expression modulo $10$ to $3(1-1+1)+7(1+1) - 0 = 17 \equiv 7$.

You are correct, the question as stated is wrong.

Answer 3

As others have observed, the $-210$ does not serve any purpose. So I'll just work on evaluating the rest of the terms modulo $10$. All this work easily generalizes.

$$ \begin{eqnarray*}3(7^{200}+7^{202}+7^{204}) + 7(3^{200}+3^{204}) & \equiv & 3((-3)^{200}+(-3)^{202}+(-3)^{204}) \\ & & - 3(3^{200}+3^{204}) \pmod{10} \\ & \equiv & 3^{203} \pmod{10} \\ & \equiv & \big(3^4\big)^{50} \cdot 3^3 \pmod{10} \\ & \equiv & 3^3 \pmod{10} \qquad \ldots \:\: \text{since} \:\:3^4 \equiv 1\pmod{10} \\ & \equiv & 7\pmod{10}. \quad \blacksquare \end{eqnarray*} $$