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# IOQM 2021 Problem Solutions

#### IOQM 2021 - Problem 1

Let $ABCD$ be a trapezium in which $AB \parallel CD$ and $AB=3CD$. Let $E$ be the midpoint of the diagonal $BD$. If $[ABCD]= n \times [CDE]$, what is the value of $n$ ? (Here $[\Gamma]$ denotes the area of the geometrical figure $\Gamma$).

Solution:

We extend $CE$ to meet $AB$ at the point $F$.

$\angle DCE = \angle EFB$, alternate angles.

Let, $X$ and $Y$ be the feet of the perpendiculars from $E$ upon $CD$ and $AB$ respectively.

Then, $X, Y, E$ are collinear, since $AB || CD$

Then, in the $\triangle CXE$ and $\triangle EFY$,

$\angle EYF = \angle EXC = 90^{\circ}$,

$\angle DCE = \angle EFB$ alternate angles, and $CE = EF$

Thus, the triangles are congruent.

Then, $EX = EY$

Now, area of trapezium =$\frac{1}{2} \times 4CD \times XY$
=$2CD \times 2EX$

=$4EX \times CD$

area of $CDE = \frac{1}{2} \times CD \times EX$

$n = 8$

#### IOQM 2021 - Problem 2

A number $N$ in base $10$, is $503$ in base $b$ and $305$ in base $b+2$. What is the product of the digits of $N$?

Solution:

$5b^2 + 3$ (number $503$ in base $b$)= $3(b+2)^2 + 5$(number $305$ in base $b+2$)
$\Rightarrow$ $5b^2 + 3$ = $3b^2 +12b + 17$
$\Rightarrow$ $2b^2 -12b - 14$ =$0$
$\Rightarrow$ $b^2 - 6b - 7$ = $0$
$\Rightarrow$ $(b +1)(b -7)$ = $0$
$b = 7$

$(503)_7$ = $5 \times 49 + 3$ = $245 + 3$ = $248$

$N = 248$

The product of the digit $N$ = $2$ $\times 4$ $\times 8$ =$64$

#### IOQM 2021 - Problem 3

If $\sum_{k=1}^{N} \frac{2k+1}{\left(k^{2}+k\right)^{2}}=0.9999$ then determine the value of $N$.
Solution:

#### IOQM 2021 - Problem 4

Let $A B C D$ be a rectangle in which $A B+B C+C D=20$ and $A E=9$ where $E$ is the mid-point of the side $B C$. Find the area of the rectangle.
Solution:

let $AB = CD= x$ & $BE = EC= y$ , $AD = 2y$

$x+y =10$, $x^2+y^2=81$

$2xy=19$

The area of a rectangle is $19$.

#### IOQM 2021 - Problem 5

Find the number of integer solutions to $||x|-2020|<5$.
Solution:

$||x| -2020|<5$

$-5<|x| - 2020< 5$

$2015<|x| <2025$

$x$ is lying between $(-2015,-2025)$ and $(2015,2025)$

There are $9$ integer between$(-2015,-2025)$ and $9$ integer between $(2015,2025)$.

So, The total $18$ integer solution.

#### IOQM 2021 - Problem 6

What is the least positive integer by which $2^{5} \cdot 3^{6} \cdot 4^{3} \cdot 5^{3} \cdot 6^{7}$ should be multiplied so that the product is a perfect square?
Solution:

By Fundamental theorem, $n=p_{1}^{n_{1}} p_{2}^{n_{2}} \cdots p_{k}^{n_{k}}=\prod_{i=1}^{k} p_{i}^{n_{i}}$

if $n$ is a perfect square then $n_i$ is even $\forall i$

$2^{5} \cdot 3^{6} \cdot 4^{3} \cdot 5^{3} \cdot 6^{7} = 2^{18} \cdot 3^{13} \cdot 5^{3}$

$3$ and $5$ has odd power then $3 \times 5 = 15$ is the minimum multiplied to make $n$ is a perfect square.

#### IOQM 2021 - Problem 7

Let $A B C$ be a triangle with $A B=A C$. Let $D$ be a point on the segment $B C$ such that $B D=48 \frac{1}{61}$ and $D C=61$. Let $E$ be a point on $A D$ such that $C E$ is perpendicular to $A D$ and $D E=11$. Find $A E$.
Solution:

$\triangle BPD \sim \triangle DBC$
$\triangle ADG \sim \triangle BDF$
$\frac{BF}{BD} = \frac{BC}{DC}$
$\Rightarrow BF =\frac{BD \times EC}{DC}=\frac{x\times \sqrt{y^2 -z^2}}{y}$
$\Rightarrow \frac{DF}{BD} = \frac{DE}{DC}$
$\Rightarrow DF = \frac{DE \times BD}{DC} = \frac{xz}{y}$
$EF = 2+ \frac{xz}{y}$
$\frac{(x+y)^2}{y}$

$AB^2 = AC^2$
$\Rightarrow BF^2 + AF^2 = AE^2 + EC^2$
$\Rightarrow AF^2 - AE^2 = EC^2 - BF^2 = (y^2 - z^2) - \frac{x^2(y^2 - Z^2)}{y^2}$
$EF^2 +2AEEF =\frac{(y^2 - z^2)(y^2 -x^2)}{y^2}$
$\Rightarrow AE \times EF =\frac{1}{2} (\frac{(y^2 - z^2)(y^2-x^2)}{y^2} -\frac {(x+y)^2 \times z^2}{y^2})$
$AE = \frac{1}{2y^2}(\frac{\frac{(y^2 - z^2)(y^2 - x^2)-(x+y)^2 z^2)}{(x+y)\times z}}{y})$

putting the values of $x,y,z$ we get $AE = 25$

#### IOQM 2021 - Problem 8

A $5$ -digit number (in base $10$ ) has digits $k, k+1, k+2,3 k, k+3$ in that order, from left to right. If this number is $m^{2}$ for some natural number $m$, find the sum of the digits of $m$ .
Solution:

$3k \leq 9$

$k = 1,2,3$

$k =1$

$\Rightarrow n = 12334$

not a perfect square as $n = 2$ (mod $4$)

$k = 2$

$\Rightarrow n = 23465$

not a perfect square as $n =15$(mod $25$)

$k = 3$

$\Rightarrow n = 34596 = 186^2$(this is a perfect square)

$\Rightarrow m = 186$

The sum of the digit $m =1 + 8 + 6 = 15$.

#### IOQM 2021 - Problem 9

Let $A B C$ be a triangle with $A B=5, A C=4, B C=6$. The internal angle bisector of $C$ intersects the side $A B$ at $D$. Points $M$ and $N$ are taken on sides $BC$ and $AC$, respectively, such that $D M || A C$ and $D N || B C$. If $(M N)^{2}=\frac{p}{q}$ where $p$ and $q$ are relatively prime positive integers then what is the sum of the digits of $|p -q|$?
Solution:

$DMCN$ is a parallelogram
$DN||MC$ and $DM || NC$ and $DC$ bisect $\angle C$
$\angle MDC = \angle DCM = \angle CDN = \angle NCD$
$\Rightarrow DM = MC =CN =ND = x$(say)

$DMCN$ is a rhombus.

Let $\triangle ADN \sim \triangle ABC$
$\frac{AN}{AC} = \frac{DN}{BC}$
$\Rightarrow \frac {AC-NC}{AC} = \frac{DN}{BC}$
$\Rightarrow \frac{4-x}{4} = \frac{x}{6}$
$x =2.4$

by cosine law,
${MN}^2$ = $2x^2(1-cos c)$ = $\frac{126}{25}$
$\Rightarrow |p-q| =101$
In $\triangle ABC$ cos c =$\frac{BC^2+AC^2-AB^2}{2BC.AC} =\frac{9}{16}$

#### IOQM 2021 - Problem 10

Five students take a test on which any integer score from $0$ to $100$ inclusive is possible. What is the largest possible difference between the median and the mean of the scores? (The median of a set of scores is the middlemost score when the data is arranged in increasing order, It is exactly the middle score when there are an odd number of scores and it is the avarage of the two middle scores when there are an even number of scores.)
Solution:

Numbers should be $0,0,0,100,100$

Median = $0$ and Mean = $40$

Difference = $40 - 0 = 40$ ( largest difference.)

#### IOQM 2021 - Problem 11

Let $X$ = $\{-5,-4,-3,-2,-1,0,1,2,3,4,5\}$ and S=$\{(a, b) \in X \times X: x^{2}+ax+ b$ and $x^{3}+bx+ a$ have at least a common real zero}. How many elements are there in $S$?
Solution:

Suppose $\alpha$ is a common root.

$\alpha^2 + 2\alpha + b = \alpha^3 +2b + \alpha = 0$

$\Rightarrow$ $a\alpha^2 -\alpha$ = $0$

$a=0$ or $\alpha = 1$ or $\alpha = -1$.

Case 1: $a = 0$ then

$b \leq 0$

$b = -5,-4,-3,-,2,-,1,0$

So, the number of element here is $6$

Case 2: $\alpha = 1$

then $1 + a + b = 0$

$\Rightarrow$ $b = -a, -1$

$a = 4,3,2,1,0,-1,-2,-3,-4,-5$.

So, the number of element here is $10$

Case 3 : $\alpha = -1$

$1 - a + b = 0$

$a = b + 1$

$b = 4,3,2,1,0,-1,-2,-3,-4,-5$.

So, the number of element here is $10$

The case of $a = 0 , \alpha = 1,-1$ is counted $22$

The total number of elements = $6 + 10 + 1 0 -2 = 24$.

#### IOQM 2021 - Problem 12

Given a pair of concentric circles, chords $A B, B C, C D, \ldots$ of the outer circle are drawn such that they all touch the inner circle. If $\angle A B C=75^{\circ},$ how many chords can be drawn before returning to the starting point.

Solution:

Let the radius of big circle be $R$ and small circle be $r$
$XY =YZ=ZZ'$
Hence all the chords are of equal length.
So, $XY=2\sqrt {R^2 - r^2}$
which is independent of $X,Y,Z,Z'$.

If the chords can be drawn, returning to the initial point, observe by how much angle $XY$ shifts to $YZ$, i.e. $X\rightarrow Y\rightarrow Z$
In $\triangle OYX$, $\angle OYX = \angle OXY=37.5^{\circ}$
As OY bisects $\angle ZYX$
Hence, $\angle XOY=105^{\circ}$

Suppose $n$ chords can be drawn.
Every single time the chord rotates by $105^{\circ}$
Therefore, $360^{\circ}$ divides $105^{\circ} \times n$

$\Rightarrow \frac{105^{\circ} \times n}{360^{\circ}}= \frac{7n}{2n}$

So, $n=24$

#### IOQM 2021 - Problem 13

Find the sum of all positive integers $n$ for which $|2^{n}+5^{n}-65|$ is a perfect square.
Solution:

$1 \leq n \leq 3$
$\Rightarrow n=2$ has a solution $\Rightarrow m=6$
$m=4 \Rightarrow \quad m=24$

For $n \leq 5$ We will see mod 10
$2^{n}$ ends with 2 or 8 if $n$ is odd
$5^{n}-65 \equiv 0$ mod 10
$2^{n}+i^{n}-65 \equiv 2 / 8 \mathrm{mod} 10$
$\Rightarrow m^{2} \equiv 01,4,5,6,9 \mathrm{mod} 10$
$\Rightarrow N+$ solution.

$n \geq 5$ mod 10
$n=\mathrm{even} =2 k$
$m^{2}= 4k+5\left(5^{2 k-1}-13\right)$
$\left(m-2^{k}\right)$
$\left(m+2^{k}\right) = 5\left(5^{2 k-1}-13\right)$
$5^{2 k-1}-13 \equiv 12$ mod $100$
$5\left(5^{2 k-1}-13\right)=60$ mod $100$
$\left(m-2^{k}\right)\left(m+2^{k}\right)=60$ mod $100$
$= 36 \times2 = 10 \times 6$
$\left(m-2^{k}\right)(m+2 k)=60$ mod $100$
There will be two possible cases $6$ (mod $100$)
$2$ (mod $100$ )
$\Rightarrow m \equiv 8$ mod $100$
$2^{k} \equiv 2$mod $100$
$100$ does not divide $2^{k} - 2$ as $4$ divides$2^{k} - 2$
$100$ does not divide $2^{k} - 14$ as $4$ divides$2^{k} - 14$
So, the number of solution is $2+4=6$

#### IOQM 2021 - Problem 14

The product $55 \times 60 \times 65$ is written as the product of five distinct positive integers. What is the least possible value of the largest of these integers?
Solution:

$55 \times 60 \times 65 = 11\times 5^3 \times 2^2 \times 3 \times 13\times 1$

$11$ and $13$ should be taken as factor because any multiple of $11$ and $13$ will give us a bigger factor.

So, we can easily see the only way to write the given expression as a product of $5$ factor where we get the minimum value of the largest factor is the following

$13 \times 11 \times 15 \times 20 \times 5$

#### IOQM 2021 - Problem 15

Three couples sit for a photograph in $2$ rows of three people each such that no couple is sitting in the same row next to each other or in the same column one behind the other. How many arrangements are possible?
Solution:

$M_1,F_1$ and $M_2,F_2$ and $M_3,F_3$ are $3$ couples.
There are $2$ cases.
Case 1: All three in a row are boys or girls.
case 2: $2$ boys and $1$ girl in one of the rows.

Case1.
$M$ = No. of arrangements in a row = $3!$
$N$ = No. of arrangements in the other row = Derangement(3) = $2$
$P$ = No of options for the row = $2$.

Total Number of arrangements in Case 1 = $MNP = 24$.

Case2.
$M$ = No. of arrangements in a row = $3! \times 3$
$N$ = No. of arrangements in the other row = Derangement(3) = $2$
$P$ = No of options for the row = $2$.

Total Number of arrangements in Case 2 = $MNP = 72$.

Total Number of arrangements = $24 + 72 = 96$

#### IOQM 2021 - Problem 16

The sides $x$ and $y$ of a scalene triangle satisfy $x+\frac{2 \Delta}{x}=y+\frac{2 \Delta}{y},$ where $\Delta$ is the area of the triangle. If $x=60, y=63$, what is the length of the largest side of the triangle?
Solution:

We obtain $\Delta$ = $\frac{xy}{2}$ where $x =60$ , $y = 63$

If, $\theta$ is angle between sides $x$, $y$ then $\Delta$ = $90^{\circ}$

Suppose $z$ is the third side

$z= \sqrt {x^2 + y^2}$ = $87$

#### IOQM 2021 - Problem 17

How many two digit numbers have exactly $4$ positive factors? (Here $1$ and the number $n$ are also considered as factors of $n$.)
Solution:

$n$ is of the form :

$n = p_1^3$

$n = p_1 . p_2$

$p_1<p_2$ are primes.

case 1 : $n = p_1^3$

only $n = 8$

So, $1$ solution here.

Case 2 : $n = p_1. p_2$

First few primes:

$2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47$

For $p_1 = 2$,

$p_2 = 5 , 7, \cdots 47$

here $13$ solutions here.

For $p_1 = 3$,

For $p_2 = 5 , 7, \cdots 31$

here $9$ solutions here.

For $p_1 = 5$,

$p_2 = 7, 11, \cdots 19$

$5$ solution here.

For $p_1 =7$,

$p_2 = 11,13$

$2$ solution here.

Total solution is $13 + 9 + 5 +2 = 29$

Thus $1 + 29 = 30$.

#### IOQM 2021 - Problem 18

If $\sum_{k=1}^{40}(\sqrt{1+\frac{1}{k^{2}}+\frac{1}{(k+1)^{2}}})=a+\frac{b}{c}$ where $a, b, c \in \mathbb{N}, b<c, gcd(b, c)=1$, then what is the value of $a+b$ ?
Solution:

$\sqrt{1 + \frac{1}{k^{2}} + \frac{1}{(k + 1)^{2}} }$

=$\sqrt\frac{k^4 + 2k^3 + 3k^2 + 2k + 1}{(k(k+1))^2}$

=$\sqrt\frac{(k^2+k+1)^2}{(k(k+1))^2}$

=$\frac{k^2+k+1}{k(k+1)}$

=$1+(\frac{1}{k} - \frac{1}{k+1})$

$\sum_{k=1}^{40}(\sqrt{1+\frac{1}{k^{2}}+\frac{1}{(k+1)^{2}}}) = 1+(\frac{1}{k} - \frac{1}{k+1})$

= $40 + (1-\frac{1}{41}) = 40 + \frac{40}{41}$

$a = 40, b = 40, c = 41$

The value of $a + b = 80$

#### IOQM 2021 - Problem 19

Let $A B C D$ be a parallelogram . Let $E$ and $F$ be midpoints of $A B$ and $B C$ respectively. The lines $E C$ and $F D$ intersect in $P$ and form four triangles $A P B$, $B P C$, $C P D$ and $D P A$. If the area of the parallelogram is $100 \mathrm{sq}$. units. what is the maximum area in sq. units of a triangle among these four triangles?
Solution:

$\triangle Q C D$
$A E || CD$
$\quad \& A E=\frac{1}{2} C D$
$\Rightarrow$ By Midpoint Theorem,
$QA=AD \ QD=2 A D$

$\triangle M P C \sim \triangle$ QPN
$\frac{P N}{P M}=\frac{Q D}{F C}=4$
$\Rightarrow P M+P N=M N=5 M P$

$[B P C]=\frac{1}{2} \times B C \times M P$
$[A P D]=\frac{1}{2} \times P N \times A D$
$\Rightarrow \frac{[B P]}{[A P D]}=\frac{M P}{P N}=\frac{1}{4}$
Let the areas $[E B P]=[EPA]=y$ Since E P is median.
$[B P F]=[F P C]=x$ $PF$ is a median
$[A P D]=a, [P D C]=b$

$[E B C]=\frac{1}{4} \times[A B C D]=2 5=2 x+y$
$[F D C]=\frac{1}{4} \times[A B C D]=2 5=2 x+b$
$[A B C D]= 100 = 2x +2y +a +b$
Solving we get $a = 25 + 3x$
$\frac{[B P C]}{[A P D]} = \frac {2x}{a} = \frac 14$
$\Rightarrow a=8 x$

$[B P C]=10$
$[A B P]=15$
$[A P D]=40$
$[P C D]=35$

#### IOQM 2021 - Problem 20

A group of women working together at the same rate can build a wall in $45$ hours. When the work started, all the women did not start working together. They joined the work over a period of time, one by one, at equal intervals. Once at work, each one stayed till the work was complete. If the first woman worked $5$ times as many hours as the last woman, for how many hours did the first woman work?
Solution:

Let there are 'n' women
$\Rightarrow$ Each woman's one hour work $=\frac{1}{45 \mathrm{n}}$
Also, $5[t-(n-1) d]=t$
$\Rightarrow \quad 4 t=5(n-1) d$
$\Rightarrow \quad \frac{1}{45 n}\left(\frac{n}{2}\right)[2 t-(n-1) d]=1$
$\Rightarrow \quad \frac{1}{90}\left[2 \mathrm{t}-\frac{4 \mathrm{t}}{5}\right]=1$
$\Rightarrow \quad t=75$ hours

#### IOQM 2021 - Problem 21

A total fixed amount of $N$ thousand rupees is given to three persons $A$. $B$. $C$ every year, each being given an amount proportional to her age. In the first year, $A$ got half the total amount. When the sixth payment was made. A got six-seventh of the amount that she had in the first year; $B$ got Rs. $1000$ less than that she had in the first year; and $C$ got twice of that she had in the first year. Find $N$.
Solution:

For A

Age at beginning = a

Money at first year= $\frac{N}{2}$

Age at $6$th payment = $a+5$

Money recieved = $\frac{6}{7}\left(\frac{\mathrm{N}}{2}\right)=\frac{3 \mathrm{~N}}{7}$

#### IOQM 2021 - Problem 22

In triangle $A B C$, let $P$ and $R$ be the feet of the perpendiculars from $A$ onto the external and internal bisectors of $\angle A B C$, respectively; and let $Q$ and $S$ be the feet of the perpendiculars from $A$ onto the internal and external bisectors of $\angle A C B$ respectively. If $P Q=7, Q R=6$ and $R S=8$, what is the area of triangle ABC?
Solution :

Let us start by drawing a picture.

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