Solve for $n$, where $n$ is a positive integer

Question

I have

$$ {n \choose 2} = 21 $$

and as the title mentions I have to solve for $n$, but so far all I have managed to get to is

$$n^2 -n =42 $$

and from there I'm completely lost. Any hints would be greatly appreciated.

Answer 1

Note that $$n^2-n=42\iff n^2-n-42=0\iff (n-7)(n+6)=0$$ and that $n\ge 2$.

Answer 2

Hint : $n^2 - n - 42 = (n-7)(n+6)$

Answer 3

$$n^2-n=(n-1)n.$$ You need to find two consecutive integers with product $42$.

Check among $2,6,12,20,30,42,56,72,90...$.

Hint: the nearest perfect squares are $6^2=36$ and $7^2=49$.

Answer 4

Another approach, if you know the formula for the sum of the first $n-1$ integers, which is $$ \sum_{k=1}^{n-1}k=\binom{n}{2} $$ We can check at each step until: $1+2+3+4+5+6=21$, and we get $n=7$.