What will be remainder when $(67^{67} + 67)$ is divided by $68$ ?

1

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63

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66

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67

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Solution

The correct option is C $66$
$(x^{n} + 1)$ will be divisible by $(x + 1)$ only when $n$ is odd.
$∴$ $(67^{67} + 1)$ will be divisible by $(67 + 1)$
$∴$ $(67^{67} + 1) + 66$ , when divided by $68$ will give $66$ as remainder.

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What will be remainder when (6767+67) is divided by 68?

The correct option is C 66(xn+1) will be divisible by (x+1) only when n is odd.∴ (6767+1) will be divisible by (67+1)∴ (6767+1)+66, when divided by 68 will give 66 as remainder.

What will be remainder when $(67^{67} + 67)$ is divided by $68$ ?

The correct option is C $66$
$(x^{n} + 1)$ will be divisible by $(x + 1)$ only when $n$ is odd.
$∴$ $(67^{67} + 1)$ will be divisible by $(67 + 1)$
$∴$ $(67^{67} + 1) + 66$ , when divided by $68$ will give $66$ as remainder.