Question

In how many ways can the number $(9!{)}^2$ be written as a sum of two or more consecutive positive integers?

• A. 19
• B. 80
• C. 20
• D. 81

Answer 1

The correct option is B

80

No. of ways of writing a number as a sum of two or more consecutive positive integers = No. of odd factors of the number - 1

$(9!{)}^2=2^{14}\times 3^8\times 5^2\times 7^2$

(No. of odd factors - 1) = 81 - 1 = 80.