Let `n` be an odd natural number greater than 1. Then , find the number of zeros at the end of the sum `99^n+1.`

The number of zeros at the end of 99^(100) - 1 is -

Number of zeroes at the end of 99^(1001)+1?

If n is an odd number greater than 1, then n(n^(2)-1) is

A natural number is greater than the other by 5. The sum of their squares is 73. find the numbers.

If n is an odd natural number greater than 1 then the product of its successor and predecessor is an odd natural number (b) is an even natural number can be even or odd (d) None of these

The sum of first n odd natural numbers is

Let n be the smallest nonprime number greater than 1 with no prime factor less than 10. Then

let 'n' be any odd natural number. Choose the incorrect statement about the number n(n^(2)-1)