If the fractional part of the number `(2^(403))/(15)` is `(k)/(15)` then k is equal to

Options

A. 14
B. 6
C. 4
D. 8

Explanation

Consider , `2^(403)=2^(400+3)=8*2^(400)=8*2^(400)=8*(2^(4))^(100) =8*(16)^(100)=8(1+15)^(100) =8(1+""^(100)C_(1)(15)+""^(100)C_(2)(15)^(2)+....+""^(100)C_(100)(15)^(100))`
[By binomial theorem , `(1+x)^(n) =""^(n)C_(0)+""^(n)C_(1)x+""^(n)C_(2)x^(2)+...""^(n)C_(n)x^(n) , n in N ]`
`=8+8(""^(100)C_(1)(15)+""^(100)C_(2)(15)^(2)+...+""^(100)C_(100)(15)^(100))`
`=8+8xx 15 lambda `
where `lambda=""^(100)C_(1)+....+""^(100)C_(100)(15)^(99)in N `
`:. (2^(403))/(15)=(8+8xx 15 lamda)/(15)=8lambda+(8)/(15)`
`implies {(2^(403))/(15)}=(8)/(15)`
(where `{*}` is the fractional part function )
`:. k=8`
Alternate Method
`2^(403)=8.2^(400)=8(16)^(100)`.
Note that, when 16 is divided by 15, gives remainder 1.
`therefore` When `(16)^(100)` is divided by 15, gives remainder `1^(100)=1` and when `8(16)^(100)` is divided by 15, gives remainder 8.
`therefore {(2^(403))/(15)}=(8)/(15)`.
(where `{.}` is the fractional part function)
`rArr k=8