Cheenta has planed to initiate a problem solving Marathon with existing students. Here we are providing the problems and hints of “Problem Solving Marathon Week 1“. The Set comprises three levels of questions as following-Level 0- for Class III-V; Level 1- for Class VI-VIII; Level 2- for the class IX-XII. You can post your alternative idea/solution in here.

## Level 0

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

Hint 1
Calculate ${1+\frac{1}{1+1}}$

Hint 2
Try to use the fact ${\frac{\frac{a}{b}}{\frac{c}{d}} =\frac{a \times d}{b \times c}}$

(Q.2)Let ${X}$ ${X}$ and ${Y}$ ${Y}$ be the following sums of arithmetic sequences:  What is the value of ${Y - X}$ ${Y - X}$?

Hint 1
Observe that how many terms are there, in $X$ and $Y$. If there are $n$ nos. of terms then pair up like $(1^{st}$ term, $n^{th}$ term $)$, $(2^{nd}$ term, ${n-1}^{th}$ term $)$

Hint 2
If you add the elements of every pair, then you will get same result of every pair.

Also Visit: Pre-Olympiad Program

## Level 1

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

Hint 1 $n^2+1$ can be written as $n(n+1)-(n-1)$

Hint 2
Try to find the necessary condition for $(n+1)|(n-1)$

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

Hint 1
Try to find the $n$th term of geometric sequence

Analyze the example $2,4.8,16$ is an example of geometric sequence. Here common ratio is $2$. Now $1^{st}$ term $=2$. $2^{nd}$ term $=1^{st}$ term $\cdot$ $2$, $3^{rd}$ term $=1^{st}$ term $\cdot$ $2^2$, $4^{th}$ term $=1^{st}$ term $\cdot$ $2^3$.

## Level 2

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

Hint 1
Note that $(m+n+p)^2 = m^2 + n^2 + p^2 + 2(mn+np+pmx)$

Hint 2
Here you can use the idea of Cauchy Schwarz Inequality