n! as sum of n divisors, n bigger than 3

Posted on January 27, 2015 by Sumant Sumant

Prove that, for each n \ge 3, the number n! can be represented as the sum of n distinct divisors of itself.

Note for n= 2, 2! it does not work because the divisors 1+2 \ne 2! but it does work for 3! where the divisors are 1+2+3. Similarly for 4! the divisors are 12+6+4+2 or 12+8+3+1.

Inductive step: Assume it is true for n = k ie. k! = d_1+d_2+...+d_k (where d_1,d_2,...d_k are k divisors in increasing order); namely multiplying the above relation by k+1 we get

(k+1)! = (k+1)d_1+(k+1)d_2+...+(k+1)d_k

(k+1)! = d_1+kd_1+(k+1)d_2+...+(k+1)d_k.

We split the (k+1)d_1 into d_1+kd_1, thus getting k+1 summands as needed. Of them kd_1 might not be a divisor of (k+1)!, that’s why we insist d_1 = 1

 

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