# Question 24c58

Feb 19, 2017

$41 \times 41 = 1681 \mathmr{and} 41 + 41 = 82$

#### Explanation:

Factors of a number are always in pairs.

The further apart the factors are, the greater is their sum and their difference. The closer they are, the less is their sum and their difference.

Consider $36$:
The factor pairs and the sum and difference are as follows:

$\text{Factors" color(white)(...........)"sum"color(white)(...........)"difference}$

$1 \times 36 \textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots} 37 \textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots . .} 35$
$2 \times 18 \textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots} 20 \textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots . .} 16$
$3 \times 12 \textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots} 15 \textcolor{w h i t e}{\ldots \ldots l . . \times x . .} 9$
$4 \times 9 \textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots .} 13 \textcolor{w h i t e}{\ldots \ldots l . . \times x . .} 5$
$6 \times 6 \textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots .} 12 \textcolor{w h i t e}{\ldots \ldots l . . \times x . .} 0$

Notice that as the factors get closer to the square root, the sum and the difference decrease.

sqrt1681 =41" rArr41 xx 41 = 1681#

$41 + 41 = 82$

This will be the smallest sum of any factors.

Note that the largest sum $= 1681 + 1 = 1682$