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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...

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  • $\begingroup$ What exactly, is meant by $\overline{aaa}$? $\endgroup$
    – user418131
    Commented Oct 1, 2017 at 5:10
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    $\begingroup$ A 3-digit number $\endgroup$ Commented Oct 1, 2017 at 5:10
  • $\begingroup$ Like $111 \times a$? $\endgroup$ Commented Oct 9, 2017 at 8:21

2 Answers 2

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$100a+10a+a=n(n+1)/2$ $\implies n^2+n=222a$ $\implies a(n/a)^2+(n/a)=222$ $\implies 2n=-1+\sqrt{1+888a}$ (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 $n$

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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 $a\in\{1,...,9\}$.

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  • $\begingroup$ Yes,it's clearly seen.But what about the given choices? $\endgroup$ Commented Oct 1, 2017 at 5:15
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    $\begingroup$ @Hamid Reza Ebrahimi The case $n=37$ does not help, but the case $n+1=37$ gives number $6$. $\endgroup$ Commented Oct 1, 2017 at 5:18
  • $\begingroup$ Can we say it decisively?! $\endgroup$ Commented Oct 1, 2017 at 5:19
  • $\begingroup$ @Hamid Reza Ebrahimi No, of course. There is only a possibility that $6$ divides $n$, $\endgroup$ Commented Oct 1, 2017 at 5:20
  • $\begingroup$ If $n=37$, then non of the given choices is acceptable,so $n+1=37$.Now the answer is CERTAINLY 6.Thank you. $\endgroup$ Commented Oct 1, 2017 at 5:23

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