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Find the value of {3^(2003)//28}, w h e ...

Find the value of `{3^(2003)//28}, w h e r e{dot}` denotes the fractional part.

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The correct Answer is:
`19//28`
`E = 3^(2003) = 3^(2001) xx 3^(2) = 9(27)^(667) = 9(28-1)^(667)`
`rArr E = 9[.^(667)C_(0) 28^(667) - .^(667)C_(1)(28)^(666) + "….."-.^(667)C_(667)]`
`= 9 xx 28k - 9`
`rArr E/28 = 9k - (9)/(28) = 9k = 1 + 19/28`
That means if we divide `3^(2003)` by `28`, the remainder is `19`. Thus, `{(3^(2003))/(28)} = 19/28`
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