# Suppose that $1+2+…+n=\overline{aaa}$. Which of the following items CERTAINLY divides $n$? $5,6,7,8,11$

Suppose that $1+2+...+n=\overline{aaa}$. Which of the following items CERTAINLY divides $n$? $5,6,7,8,11$

I converted the given relation into the following: $$n(n+1)=2*3*37*a$$
Now I think we must consider all different cases of divisibility but can't give reasoning...

• What exactly, is meant by $\overline{aaa}$? – AnotherJohnDoe Oct 1 '17 at 5:10
• A 3-digit number – Hamid Reza Ebrahimi Oct 1 '17 at 5:10
• Like $111 \times a$? – steven gregory Oct 9 '17 at 8:21

From you equality you can get that $n=37$ or $n+1=37$ because

even if $n+1=74$ you can not get the number $100a+10a+a$, where $a\in\{1,...,9\}$.

• Yes,it's clearly seen.But what about the given choices? – Hamid Reza Ebrahimi Oct 1 '17 at 5:15
• @Hamid Reza Ebrahimi The case $n=37$ does not help, but the case $n+1=37$ gives number $6$. – Michael Rozenberg Oct 1 '17 at 5:18
• Can we say it decisively?! – Hamid Reza Ebrahimi Oct 1 '17 at 5:19
• @Hamid Reza Ebrahimi No, of course. There is only a possibility that $6$ divides $n$, – Michael Rozenberg Oct 1 '17 at 5:20
• If $n=37$, then non of the given choices is acceptable,so $n+1=37$.Now the answer is CERTAINLY 6.Thank you. – Hamid Reza Ebrahimi Oct 1 '17 at 5:23

$100a+10a+a=n(n+1)/2$ $\implies n^2+n=222a$ $\implies a(n/a)^2+(n/a)=222$ $\implies 2n=-1+\sqrt{1+888a}$ (Negative sqrt is rejected) As $n$ is a natural number, so is $2n$. The quantity under the radical must be a perfect square and its square root must be greater than 1.

Again, as $a \in \left\{1,2,3,5,6,7,8,9\right\}$ The quantity $\sqrt{1+888a}$ is a positive integer only for $a=6$. This gives $n=36$

Hence, among the choices given, $6$ perfectly divides $n$