 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.