There are three consecutive positive integers such that the sum of the squares of the smallest two is 221. What are the numbers?

1 Answer
Jun 22, 2016

There are 10, 11, 12.

Explanation:

We can call the first number n. The second number has to be consecutive, so it will be n+1 and the third one is n+2.

The condition given here is that the square of the first number n^2 plus the square of the following number (n+1)^2 is 221. We can write

n^2+(n+1)^2=221

n^2+n^2+2n+1=221

2n^2+2n=220

n^2+n=110

Now we have two methods to solve this equation. One more mechanics, one more artistic.
The mechanics is to solve the second order equation n^2+n-110=0 applying the formula for the second order equations.
The artistic way is to write

n(n+1)=110

and observe that we want that the product of two consecutive numbers has to be 110. Because the numbers are integer we can search these numbers in the factors of 110. How can we write 110?

For example we notice that we can write it as 110=10*11.

Oh, it seems we found our consecutive numbers!

n(n+1)=10*11.

Then n=10, n+1=11 and, the third number (not very useful for the problem) n+2=12.