BdMO National Secondary 2007#7
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- Posts:1007
- Joined:Sat Dec 09, 2017 1:32 pm
Find the smallest positive integer $n>1$, such that $\sqrt{1+2+3+...+n}$ is an integer.$(n<10)$.
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- Posts:1007
- Joined:Sat Dec 09, 2017 1:32 pm
Re: BdMO National Secondary 2007#7
Answer: $\fbox 8$
$\sqrt{1+2+3+4+5+6+7+8}\Rightarrow \sqrt{36}\Rightarrow 6$
And $1<8<10$.
So our desired $n=8$.
$\sqrt{1+2+3+4+5+6+7+8}\Rightarrow \sqrt{36}\Rightarrow 6$
And $1<8<10$.
So our desired $n=8$.