This is a hyperbola. It is best approached using Fermat's factorisation method. See

fermat-s-factorization-method

or google wikepedia. I don't know of any faster approach.

This is a hyperbola. It is best approached using Fermat's factorisation method. See

fermat-s-factorization-method

or google wikepedia

Use Fermat Factorization to work out this factorization!

Note that 115 < √13431 < 116.

Select the squared value of a number, say 116. Then:

116² - 13431 = 25

Since 25 is the perfect square number, 116 works. Now, using this form:

n = s² - t² = (s - t)(s + t) where t is the value of the square root of some perfect square

We obtain:

13431 = (116 - 5)(116 + 5)

= 111(121)

Now, this should look obvious. Factor out each term by term to get:

111 x 121

= 3 x 37 x 11² or 3 x 11² x 37

So there are some numbers that are divisible by 13431. They are:

1. 3
2. 11
3. 37
4. 121
5. 111
6. 1221
7. 4477
8. 407

Note: Fermat Factorization only applies to factorizable odd number!