Find the remainder when $9^{400} + 9^{401} + 9^{402} . . . . . . . . . . 9^{410}$ is divided by 6?

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Solution

The correct option is C
3

Any power of 9 when divided by 6, gives a remainder 3. hence answer $= \frac{11 \times 3}{6}$ ((As there are 11 terms) $= \frac{33}{6}$ = remainder =3

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