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The remainder when 5^(99) is divided by ...

The remainder when `5^(99)` is divided by 13 is

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Here, `5^(2) = 25 = 26-1`
So, `5^(99) = 5 xx 5^(98) = 5(5^(2))^(49) = 5(26-1)^(49)`
`= 5[.^(49)C_(0)26^(49)-.^(49)C_(1)26^(48)+.^(49)C_(2)26^(47)-"….."+.^(49)C_(48)26-.^(49)C_(49)]`
`= 5(26k-1)`
`= 130k - 5`
`:. (5^(99))/(13) = (130 k - 5)/(13) = (130 k -13 + 8)/(13) = (10 k - 1) + 8/13`
Hence, the remainder is `8`.
Clearly, intergral part `(10k - 1)` or `(5^(99))/(13)` is an odd number.
Also , fractional part of `(5^(99))/(13)` is `(8)/(13)`.
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