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 October 5th, 2019, 03:46 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 Probability involving square numbers Let $\displaystyle n\leq 111 \:$ , $\displaystyle n\in \mathbb{N}$. Find the probability such that n is a square-number and odd . $\displaystyle P(n=N^{2} )$=? Last edited by idontknow; October 5th, 2019 at 04:09 AM.
October 5th, 2019, 04:26 AM   #2
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Quote:
 Originally Posted by idontknow Let $\displaystyle n\leq 111 \:$ , $\displaystyle n\in \mathbb{N}$. Find the probability such that n is a square-number and odd . $\displaystyle P(n=N^{2} \& n=1(mod 2) )$=?
All square numbers below 111:
$\sqrt{111} = 10.53565375285273884840140466189966747674770065972 434833044...$, therefore biggest square number below 111 is $100=10^2$. Therefore, there are 10 square number below 111.

1,4,9,16,25,36,49,64,81,100. There 10 square numbers below 111. And five of them is odd.

Therefore, $P(n=N^{2} \; \& \; n \equiv 1(mod 2) ) = 5/111$

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