Question

the remainder when 2^59! is divided by 255 is

Answer 1

Now 59! has 8 as a factor, since 59 ! = 59 * 58 * 57 * …..* 8 * 7 * ….* 1.

Let us write 59! as 8* k, where k is a (large) integer.

Then 259!=(28)k

= 256k

= (255+1)k

Using binomial expansion on above, we have (255+1)k

= Multiple of 255 + 1.

Thus 259! = Multiple of 255 + 1.

So, 259!(mod255)=1 .

1 is the required answer.

OR

It is very easy to verify that 59! is divisible by 128.

59!=k×128

259!(mod255)

=2k×128(mod255)

=(2128)k(mod255)

=(2ϕ(255))k(mod255) [ ϕ(255)=128 ( ϕ is the )]

=1k [By we get 2ϕ(255)≡1(mod255) ]

=1