Question

What is the largest number of zeroes that can occur at the end of `1^n+2^n+3^n+4^n` for any positive integer n?

Answer 1

`2`

Explanation:

`1^n + 2^n + 3^n + 4^n = k*10^p`
`=> 2^n + 2^(2*n) = k*10^p - 3^n - 1`
`=> 2^n * (1 + 2^n) = k*2^p*5^p - 3^n - 1`

`"Now, assuming that "n >= p", we can say that"`

`"LHS divisible by "2^n => " RHS divisible by "2^n`
`=> " RHS divisible by "2^p`

`"As "k*2^p*5^p" is already divisible by "2^p", we have the"`
`"condition that"`

`3^n + 1 " is divisible by "2^p`

`"Now take "p=3`

`3^n + 1 " is NOT divisible by "2^3 = 8`

`"We can prove this as follows"`

`3^n mod 8 != 5 and`
`3^n mod 8 != 7" (induction hypothesis)"`
`=> 3^(n+1) mod 8 = (3 * (3^n mod 8)) mod 8`
`"So we assume that "(3^n mod 8)" can only be 0,1,2,3,4,6."`
`"Then"`
`(3 * 0) mod 8 = 0`
`(3 * 1) mod 8 = 3`
`(3 * 2) mod 8 = 6`
`(3 * 3) mod 8 = 1`
`(3 * 4) mod 8 = 4`
`(3 * 6) mod 8 = 2`
`"So the same goes for "3^(n+1) mod 8.`
`"We conclude that"`
`3^n mod 8 != 7 => 3^n + 1 " is NOT divisible by 8."`

`"So p cannot be bigger than 2."`