Problem Solving Marathon Week 2

We are having a full fledged Problem Solving Marathon. We are receiving wonderful responses from the end of our students which is making the session more and more alluring day by day. Here we are providing the problems and hints of "Problem Solving Marathon Week 2". 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] In triangle $latex CAT$, we have $latex \angle ACT =\angle ATC$ and $latex \angle CAT = 36^\circ$. If $latex \overline{TR}$ bisects $latex \angle ATC$, then $latex \angle CRT =$

Hint 1
Triangle $latex CAT$ is an isosceles triangle

Hint 2
Try to find the value of $latex \angle ATC$

[Q.2] What is the product of $latex \frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\cdots\times\frac{2019}{2018}$?

Hint 1
Figure out the hidden pattern of the problem.

Hint 2
You can make a prototype of that problem i.e, try to calculate product of first three or four terms.

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Level 1

[Q.1] How many $latex 7$-digit palindromes (numbers that read the same backward as forward) can be formed using the digits $latex 2$, $latex 2$, $latex 3$, $latex 3$, $latex 5$, $latex 5$, $latex 5$?

Hint 1
First symbolize the number i.e., the number looks like "$latex \textbf{mnoponm}$".

Hint 2
Now $latex \textbf{p}$ cannot be $latex 2$ or $latex 3$.

[Q.2] How many pairs of positive integers $latex (a,b)$ are there such that $latex a$ and $latex b$ have no common factors greater than 1 and: $latex \frac{a}{b} + \frac{14b}{9a}$ is an integer?

Hint 1
put $latex x=\frac{a}{b}$

Hint 2
Discriminant of a Quadratic equation plays a significant role here..

Level 2

[Q.1] Prove that if $latex m$, $latex n$ are integers, then the expression $latex E = m^5 + 3m^4n - 5m^3n^2 - 15m^2n^3 + 4mn^4 + 12n^5$ cannot take the value $latex 33$.

Hint 1
Factorise the given expression..

Hint 2
Then try to find the pairwise different divisors $latex 33$

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 $latex 1 + \frac{1}{1+\frac{1}{1+1}}$?

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.

Problem Solving Marathon Week 1

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 $latex {1 + \frac{1}{1+\frac{1}{1+1}}}$?

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

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

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

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

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

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Level 1

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

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

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

(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 $latex 2$.

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

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

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 $latex (1+m)(1+n)(1+p) \leq 4 + 4mnp$

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

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