   SEARCH HOME Math Central Quandaries & Queries  Question from Kenneth: Hello: Is there a fast and simple method that can be used to determine the aliquot parts, whole, positive numbers that can divide into another whole, positive number? For example, the aliquot parts for 10 are 1, 2, 5, 10. One method is to divide the number by whole numbers from 1 up to and including that whole number. If the number is large, for example 1000, the division would be long and tiresome. I thank you for your reply. Hi Kenneth,

You taught me a term I didn't know, aliquot parts. I would use the term divisor.

Definition:
If $a$ and $c$ are positive integers then $a$ is a divisor of $c$ if there is an integer $b$ so that $a \times b = c.$

The short answer to your question is no. There is a reasonable looking expression for the number of divisors of a positive integer but it is impractical. Let me look at an example.

Find the divisors (aliquot parts) of 2,045,407.

First find the prime factorization of 2,045.407. Without much trouble you can see that

$7^2 \times 13^3 \times 19 = 2,045,407$

Any divisor is then a product of 0,1 or 2 sevens, times a product of 0, 1 ,2 or 3 thirteens, times a product of 0 or 1 nineteen. Hence there are

$(2+1) \times (3 + 1) \times (1+1) = 24 \mbox{ divisors of 2,045,407.}$

With a few minutes of time you can write out all 24 divisors.

The problem with this method is that you first need to write the prime factorization of the number. If the number has a few million digits there is no practical way to find its prime factorization. To factor large numbers is extremely difficult, if not impossible, even for the world's fastest computers.

Harley     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.