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# Problem Solving Marathon Week1 Solution

Problem Solving Marathon Week1 Solution is the effortless attempt from Cheenta's existing student as well as from the end of mentor. Question, rules and hints are given here.

## Level 0

Q.1 Which of the following is equal to $1 + \frac{1}{1+\frac{1}{1+1}}$?

This solution is proposed by Swetaabh Mishra from Thousand Flowers. $1 + \frac{1}{1+\frac{1}{1+1}} = 1 + \frac{1}{1+\frac{1}{2}}$ $=1 + \frac{1}{\frac{3}{2}}= 1+\frac{2}{3}=1\frac{2}{3}$

Answer of (Q.2) This solution is using hints.
If we pair up the elements of $X$ it will look like $(10,100)(12,98),(14,96),.....(54,56)$. Now sum of the each pair is $110$. Number of pair $=\frac{Number of terms}{2} =\frac{46}{2}$. so $X$ will be equal to Number of pair $\times 110 =\frac{46}{2} \times 110=2530$ , similarly $Y$ will be $\frac{46}{2} \times 114=2622$.
So, $Y-X$ will be $92$.

## Level 1

Q.1 Find all positive integers $n$ such that $n^2+1$ is divisible by $n+1$.

Since, $n^2+1$ can be written as $n(n+1)-(n-1)$
We can say that if $n+1|n^2+1$ then $n+1|n-1$. At a glance, it's look like impossible to get any positive integer. If $n-1=0$ then it is possible. So there is only one such positive integer $n=1$.

Q.2 Two geometric sequences $a_1, a_2, a_3, \ldots$ and $b_1, b_2, b_3, \ldots$ have the same common ratio, with $a_1 = 27$, $b_1=99$, and $a_{15}=b_{11}$. Find $a_9$.
Example of Geometric Sequence $2,4,8,16$, here common ratio is $2$.

This solution is proposed by Saikrish Kailash from Thousand Flowers.
Two geometric sequence have same common ratio, let it is $r$.
Now $a_9=27\times r^{(9-1)}=27r^8$. From the given condition $a_{15}=b_{11}$. Which is equivalent to $27r^{14}=99r^{10} \Rightarrow r^4=\frac{11}{3}$. Now put the value of $r^4$ in $a_9=27r^8=27(r^4)^2=363$.

## Level 2

Q.1 Let $m$, $n$, $p$ be real numbers such that $m^2 + n^2 + p^2 - 2mnp = 1$ . Prove that $(1+m)(1+n)(1+p) \leq 4 + 4mnp$.

This solution is proposed by Sampreety Pillai from Early Bird Math Olypmiad Group.
We konw $(m-n)^2 \geq 0$ which imply $m^2+n^2 \geq 2mn$. similarly $n^2+p^2 \geq 2np$, $p^2+m^2 \geq 2mp$. Also $m^2+1 \geq 2m$, $n^2+1 \geq 2n$ and $p^2+1 \geq 2p$. Adding these six inequalities, we get $3(m^2+n^2+p^2+1) \geq 2(mn+mp+np+m+n+p)$. by the hypothesis $m^2+n^2+p^2=2mnp+1$.
So $3(mnp+1) \geq 2(mn+mp+np+m+n+p)$. Add $mnp+1$ both side of last inequality.
Also you can do another type of proof using Cauchy-Schwarz inequality. For that click here.

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