The remainder when $1^{1997} + 2^{1997} + . . . . . . + 1996^{1997}$ is divded by $1997$ is

0

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1

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197

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1996

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Solution

The correct option is D $0$
According to fermat's Little Theorem,
. Since $1997$ is a prime number,
$∴ 1^{1997}, 2^{1997}, . . . . ., 1996^{1997} a r e d i v i d e d b y 1997 t h e n r e m a i n d e r w i l l b e r e s p e c t i v e l y 1, 2, . . ., 1996$
$1 + 2 + . . . . . + 1996 = \frac{1}{2} \times 1996 \times 1997 = 998 \times 1997$
Which is divisible by $1997$ .
Hence remainder $= 0$

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