Convincing Sum of Divisors?

Number Theory Level 3

If the sum of all the positive divisors of the positive integer $N$ is equal to $1 + 2 + 4 + 8 + 16 = 31$ , does it necessarily imply that $N$ is equal to 16?

Yes No

3 solutions

Vilakshan Gupta
Nov 30, 2017

Let $\sigma(N)$ represent the sum of positive divisors of $n$ .

Note that for any positive integer $N$ , $\sigma(N) \ge N$ .

Therefore after calculating $\sigma(N)$ for numbers upto $31$ , we get $\sigma(25)=1+5+25=31$ . Therefore, $\sigma(16)$ is not unique to $16$ .

A simple way to justify it to first show that $N$ cannot exceed 31.

Plus, you have already shown another value that satisfies this condition, so your solution is pretty much satisfactory.

Pi Han Goh - 3 years, 6 months ago

Oh Yes. I didn't observe that. Thanks. I will change the solution.

Vilakshan Gupta - 3 years, 6 months ago

Leonel Castillo
Jan 24, 2018

Here is another, perhaps more systematic, approach:

Nice approach. I didn't thought about solving it this way. +1

Pi Han Goh - 3 years, 4 months ago

Munem Shahriar
Nov 30, 2017

When $N= 25,$ the sum of its divisors is

$1+5+25 = 31$

Hence $N = 16$ is not the only number that satisfies the condition.

Convincing Sum of Divisors?

3 solutions

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