Sum of three numbers

Algebra Level pending

The sum of $3$ distinct positive integers is $20$ . The largest of these $3$ is the sum of the $2$ smallest. How many solutions are there? If the integers are $a,b$ and $c$ , let $a<b<c$ .

4 2 5 3

1 solution

A Former Brilliant Member
Oct 2, 2017

Let $a,b$ and $c$ be the integers such that $a<b<c$ .

Therefore $a+b+c=20$ . However, $a+b=c$ . So

$c+c=20$

Thus, $c=10$ .

So the solutions are: $(10,9,1),(10,8,2),(10,7,3)$ and $(10,6,4)$ .

Hey, I believe there are 8 solutions in total..

In your solution you took $c$ as the largest integer.That is fine. But in the question nowhere does it say that from the other two integers, which one is smaller or which one is bigger.You arbitarly took $b$ is greater than $a$ . Howe ver if you remove that condition, there will be 8 solutions in total, namely $(10,9,1)$ $(10,8,2)$ $(10,7,3)$ $(10,6,4)$ $(10,4,6)$ $(10,3,7)$ $(10,2,8)$ $(10,1,9)$

Aman thegreat - 3 years, 8 months ago

Sum of three numbers

1 solution

0 pending reports