Find the remainder when `x=5^5^5^dot^5` (24 times 5) is divided by 24.

Explanation

Here , ` 5^(5^(5^(...^(5))))` (23 times 5 ) is an odd natural number
Let ` 5^(5^(5^(...^(5))))` (23 times 5) = 2m + 1
Now , let ` x = 5^(5^(5^(...^(5)))) "(24 times 5)" = 5^(2m+1) = 5 * 5^(2m) ` , where m is
a natrual number
` therefore x = 5 * (5^(2))^(m) = 5 (24 + 1)^(m)`
` = 5[""^(m)C_(0) (24)^(m)+ ""^(m)C_(1) (24)^(m-1)+ ...+ ""^(m)C_(m-1) (24) + 1]`
`= 5 (24 k + 1) = 24 (5k) + 5 `
` therefore (x)/(24) = 5k + (5)/(24)`
Hence , the remainder is 5