# The sum of two numbers is 41. One number is less than twice the other. How do you find the larger of the two numbers?

Jun 28, 2015

The conditions are not restrictive enough. Even assuming positive integers the larger number can be any number in the range $21$ to $40$.

#### Explanation:

Let the numbers be $m$ and $n$

Assume $m , n$ are positive integers and that $m < n$.

$m + n = 41 = 20.5 + 20.5$

So one of $m$ and $n$ is less than $20.5$ and the other is greater.

So if $m < n$, we must have $n \ge 21$

Also $m \ge 1$, so $n = 41 - m \le 40$

Putting these together, we get $21 \le n \le 40$

The other condition that one number is less than twice the other is always satisfied, since $m < 2 n$