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JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Binomial Theorem

The remainder on dividing

5^{99}

by 11 is

Answer: 9

Solution

The powers of

5

repeat modulo

11

:

5^{1} \equiv 5

,

5^{2} \equiv 3

,

5^{3} \equiv 4

,

5^{4} \equiv 9

and

5^{5} \equiv 1 (mod 11)

, so the pattern closes after five steps. Writing

99 = 5 \times 19 + 4

gives

5^{99} = (5^{5})^{19} \cdot 5^{4} \equiv 1^{19} \cdot 5^{4} \equiv 9 (mod 11)

. Hence division of

5^{99}

by

11

leaves the remainder

9

.

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Integer answer

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