Mininum Sum

Algebra Level 3

What is the least possible sum of two positive integers $a$ and $b$ where $a \times b = 10!$

Notation ! is a factorial symbol; for example,

$3!=3\times2\times1$

3810 3840 3900 3245 3456

1 solution

Andrew Lamoureux
Oct 5, 2017

$10! = 1*2*2*2*2*2*2*2*3*3*3*3*5*7*10$

All of these prime factors must be placed in the problem's two factors $a, b$ .

Starting from the most unbalanced placement, a trend is visible:

$1+2*2*2*2*2*2*2*3*3*3*3*5*7*10 = 3628801$

$1*2+2*2*2*2*2*2*3*3*3*3*5*7*10 = 1814402$

$1*2*2+2*2*2*2*2*3*3*3*3*5*7*10 = 907204$

$1*2*2*2+2*2*2*2*3*3*3*3*5*7*10 = 453608$

A perfect balance is given by the square root operation: $n = \sqrt{n} * \sqrt{n}$ .

But $\sqrt{10!} = 1904.941$ isn't whole. The nearest whole number factor pair is $1890*1920$ . And they sum to $3810$ .

Thank you for sharing your solution.

Hana Wehbi - 3 years, 8 months ago