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 $latex 1 + \frac{1}{1+\frac{1}{1+1}}$?
This solution is proposed by Swetaabh Mishra from Thousand Flowers.
$latex 1 + \frac{1}{1+\frac{1}{1+1}} = 1 + \frac{1}{1+\frac{1}{2}}$
$latex =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 $latex X$ it will look like $latex (10,100)(12,98),(14,96),.....(54,56)$. Now sum of the each pair is $latex 110$. Number of pair $latex =\frac{Number of terms}{2} =\frac{46}{2}$. so $latex X$ will be equal to Number of pair $latex \times 110 =\frac{46}{2} \times 110=2530$ , similarly $latex Y$ will be $latex \frac{46}{2} \times 114=2622$.
So, $latex Y-X$ will be $latex 92$.
Level 1
Q.1 Find all positive integers $latex n$ such that $latex n^2+1$ is divisible by $latex n+1$.
Since, $latex n^2+1$ can be written as $latex n(n+1)-(n-1)$
We can say that if $latex n+1|n^2+1$ then $latex n+1|n-1$. At a glance, it's look like impossible to get any positive integer. If $latex n-1=0$ then it is possible. So there is only one such positive integer $latex n=1$.
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Q.2 Two geometric sequences $latex a_1, a_2, a_3, \ldots$ and $latex b_1, b_2, b_3, \ldots$ have the same common ratio, with $latex a_1 = 27$, $latex b_1=99$, and $latex a_{15}=b_{11}$. Find $latex a_9$.
Example of Geometric Sequence $latex 2,4,8,16$, here common ratio is $latex 2$.
This solution is proposed by Saikrish Kailash from Thousand Flowers.
Two geometric sequence have same common ratio, let it is $latex r$.
Now $latex a_9=27\times r^{(9-1)}=27r^8$. From the given condition $latex a_{15}=b_{11} $. Which is equivalent to $latex 27r^{14}=99r^{10} \Rightarrow r^4=\frac{11}{3}$. Now put the value of $latex r^4$ in $latex a_9=27r^8=27(r^4)^2=363$.
Level 2
Q.1 Let $latex m$, $latex n$, $latex p$ be real numbers such that $latex m^2 + n^2 + p^2 - 2mnp = 1$ . Prove that $latex (1+m)(1+n)(1+p) \leq 4 + 4mnp$.
This solution is proposed by Sampreety Pillai from Early Bird Math Olypmiad Group.
We konw $latex (m-n)^2 \geq 0$ which imply $latex m^2+n^2 \geq 2mn$. similarly $latex n^2+p^2 \geq 2np$, $latex p^2+m^2 \geq 2mp$. Also $latex m^2+1 \geq 2m$, $latex n^2+1 \geq 2n$ and $latex p^2+1 \geq 2p$. Adding these six inequalities, we get $latex 3(m^2+n^2+p^2+1) \geq 2(mn+mp+np+m+n+p)$. by the hypothesis $latex m^2+n^2+p^2=2mnp+1$.
So $latex 3(mnp+1) \geq 2(mn+mp+np+m+n+p)$. Add $latex mnp+1$ both side of last inequality.
Also you can do another type of proof using Cauchy-Schwarz inequality. For that click here.
