Modulo

$\textbf{Hint:}$

$2^{p-1} \equiv 1 \pmod{3}$ for any odd prime p.

Since the problem asked the remainder for a perfect number $mod$ $6$ , I just found the remainder with respect to the first few perfect numbers i.e. $28, 496,$ etc. They all turned out to be $4$ .

Hence, our answer is $\boxed{4}$ .

For perfect number ..... 2^(p−1) * (2^p −1) is an even perfect number , here p=2,3....... if p=2 then perfect number = 2^1(2^2 -1) = 6 if p-3 then perfect number = 2^2(2^3 -1) = 28

here n>6 and n is even perfect number, so, n=28 and 28%6=4 so, ans=4

Hello pals,

as you all know even perfect numbers are = [ 6,28,496,8128,....]

since n > 6 , n = [28,496,8128,....] or you can use 2^(p-1)(2^p -1) , where p=[2,3,5,7,....]

i'm just taking n = 28,496 and 8128 to show the remainder when divided by six, as there are only certain perfect numbers are discovered,

n = 28 , 28 / 6 => remainder is 4

n = 496 , 496 / 6 => remainder is 4

n = 8128 , 8128 / 6 => remainder is 4

therefore, remainder is 4...

thanks....

28,496 are perfect numbers which when divided by 6 leaves the remainder 4. Since this question has only one answer, the answer must be 4 for all the remaining perfect numbers.