Divisors Identity

Number Theory Level 4

Inclusive of 1 and itself, let $d_1,d_2,\ldots,d_n$ denote the distinct positive divisors of positive integer $N$ . Given that $\displaystyle \sum_{m=1}^n d_m = 4056 \times18$ and $\displaystyle \sum_{m=1}^n \frac1{d_m} = \frac{4056}{25\times55}$ , find the value of $N$ .

The answer is 24750.

1 solution

Kartik Sharma
Nov 12, 2015

This is quite simple but neat.

Pre-requisites would be to observe :

if $d_m|N$ or $N = d_m \times d_{n-m}$ , then $\displaystyle d_{n-m+1} = \frac{N}{d_m}$

Now, we are given $\displaystyle \sum_{m=1}^{n}{\frac{1}{d_m}} = \frac{4056}{25\times 55}$

Multiply by $N$ both sides,

$\displaystyle \sum_{m=1}^{n}{\frac{N}{d_m}} = \sum_{m=1}^{n}{d_{n-m+1}} = \frac{4056\times N}{25\times 55}$

It is given that $\displaystyle \sum_{m=1}^{n}{d_m} = 4056 \times 18$

Hence, $\displaystyle N = 18 \times 25 \times 55 = 24750$

Yep. There's a bijection between $d_i$ and $\frac{N}{d_i}$ .

Pi Han Goh - 5 years, 7 months ago

Divisors Identity

The answer is 24750.

1 solution

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