What is the remainder when $9^3 + 9^4 + 9^5 + 9^6 + \ldots + 9^{100}$ is divided by 6?
What is the remainder when $9^3 + 9^4 + 9^5 + 9^6 + \ldots + 9^{100}$ is divided by 6?
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Solution
Each power of 9 (for exponent $\ge1$) leaves remainder 3 when divided by 6, since $9\equiv3\pmod6$ and $9^k\equiv3\pmod6$ for $k\ge1$. The sum runs from $9^3$ to $9^{100}$, that is $100-3+1 = 98$ terms, each $\equiv3\pmod6$. Total $\equiv98\times3 = 294\pmod6$. Since $294 = 6\times49$, the remainder is 0.