BdMO National Secondary 2007#7

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samiul_samin
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Joined:Sat Dec 09, 2017 1:32 pm
BdMO National Secondary 2007#7

Unread post by samiul_samin » Sat Feb 23, 2019 9:13 am

Find the smallest positive integer $n>1$, such that $\sqrt{1+2+3+...+n}$ is an integer.$(n<10)$.

samiul_samin
Posts:1007
Joined:Sat Dec 09, 2017 1:32 pm

Re: BdMO National Secondary 2007#7

Unread post by samiul_samin » Sat Feb 23, 2019 11:14 am

Answer: $\fbox 8$
$\sqrt{1+2+3+4+5+6+7+8}\Rightarrow \sqrt{36}\Rightarrow 6$
And $1<8<10$.
So our desired $n=8$.

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