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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:

Watch this video.

#### 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}=$4k+5\left(5^{2 k-1}-13\right)\left(m-2^{k}\right)\left(m+2^{k}\right) = 5\left(5^{2 k-1}-13\right)5^{2 k-1}-13 \equiv 12 mod 1005\left(5^{2 k-1}-13\right)=60$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?

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 $N$.

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?

Let us start by drawing a picture.

#### IOQM 2021 – Problem 23

The incircle $\Gamma$ of a scalene triangle $A B C$ touches $B C$ at $D$, $CA$ at $E$ and $A B$ at $F$. Let $r_{A}$ be the radius of the circle inside $A B C$ which is tangent to $\Gamma$ and the sides $A B$ and $A C$. Define $r_{B}$ and $r_{C}$ similarly. If $r_{A}=16, r_{B}=25$ and $r_{C}=36,$ determine the radius of $\Gamma$.
Solution:

Using the formula
$r =\sqrt{r_{a} \cdot r_{b}}+\sqrt{r_{b} \cdot r_{c}}+\sqrt{r_{c} \cdot r_{a}}$
=$\sqrt{16 \cdot 25}+\sqrt{25 \cdot 36}+\sqrt{36 \cdot 16}$
=$20+30+24=74$

#### IOQM 2021 – Problem 24

A light source at the point $(0,16)$ in the coordinate plane casts light in all directions. A disc (a circle along with its interior) of radius 2 with center at (6,10) casts a shadow on the X axis. The length of the shadow can be written in the form $m \sqrt{n}$ where $m , n$ are positive integers and $n$ is square-free. Find $m+n$.
Soulition:

MPB has slope -1
$\Rightarrow MBO=45^{\circ}$
$\angle A M B=\angle B M C=\alpha$

$\Rightarrow \angle MAO =45^{\circ} +\alpha$

$\Rightarrow \angle MCO =45^{\circ} – \alpha$

$A C=OC-OA$

$MO=16$

$P M= 6 \sqrt{2}$

$O A=\frac{16}{\tan (45)^{\circ} + \alpha)}$

$O Q=\frac{16}{\tan (45)^{\circ} – \alpha)}$

$A C=OC-OA = 16 \times 4 \frac {\tan \alpha}{1 -\tan ^ 2 \alpha}$

$\triangle MPX \tan \alpha = \frac {1}{\sqrt 7}$

PX= $2$, MP =$6 \sqrt{2}$

AC = $4 \sqrt{17}$

m+n = $4$

#### IOQM 2021 – Problem 25

For a positive integer $n$, let $(n)$ denote the perfect square integer closest to $n$. For example, $\langle 74\rangle=81,(18)=16$ If $N$ is the smallest positive integer such that $(91) \cdot(120) \cdot\langle 143\rangle \cdot\langle 180\rangle \cdot\langle N\rangle=91 \cdot 120 \cdot 143 \cdot 180 \cdot N$. Find the sum of the squares of the digits of $N$.

$100 \cdot 121 \cdot 144 \cdot 169 (N) = 91 \times \cdot(120) \cdot\langle 143\rangle \cdot\langle 180\rangle \cdot\langle N\rangle=91 \cdot 120 \cdot 143 \cdot 180 \cdot N$.

$\quad 81 \cdot 121 \cdot 144 \cdot 169 \cdot\langle\mathrm{N}\rangle=91 \cdot 120 \cdot 143 \cdot 180 \cdot \mathrm{N}$

$\Rightarrow \quad\langle\mathrm{N}\rangle=\frac{91 \cdot 120 \cdot 143 \cdot 180 \cdot \mathrm{N}}{100 \cdot 121 \cdot 144 \cdot 169}$
$\Rightarrow \quad\langle\mathrm{N}\rangle=\frac{21}{22} \mathrm{~N}$

Now to make $\langle\mathrm{N}\rangle$ to be a perfect square, we can take smallest $\mathrm{N}$ to be $2 \cdot 11 \cdot 3 \cdot 7=162$
$\quad\langle N\rangle=\frac{21}{22} N=\frac{3 \cdot 7 \cdot 2 \cdot 11 \cdot 3 \cdot 7}{2 \cdot 11}=(21)^{2}=441$
Which is the nearest perfect square to $462$.

$\quad$ Sum of square of digits of 462 is $4^{2}+6^{2}+2^{2}$
$$=16+36+4=56$$

#### IOQM 2021 – Problem 26

In the figure below, 4 of the 6 disks are to be colored black and 2 are to he colored white. Two coloring that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same.

There are only four such colorings for the given two colors, as shown in Figure $1 .$ In how many ways can we color the 6 disks such that 2 are colored black. 2 are colored white, 2 are colored blue with the given identification condition?

#### IOQM 2021 – Problem 27

A bug travels in the coordinate plane moving only along the lines that are parallel to the $x$ axis or $y$ axis. Let $A=(-3,2)$ and $B(3,-2)$. Consider all possible paths of the bug from $A$ to $B$ of length at most $14$ . How many points with integer coordinates lie on at least one of these paths?

#### IOQM 2021 – Problem 28

A natural number $n$ is said to be good if $n$ is the sum of $r$ consecutive positive integers, for some $r \geq 2$. Find the number of good numbers in the set ${1,2…… 100}$.

$k + (k+1)+ \cdots +(k+r-1) = n$

$\Rightarrow 2n = (2k + r – 1)\times r$

$n = r \times (2k + r -1)$

$\Rightarrow$ n has two factors $r$ and $2k+r-1$

Then, $- r + (2k+r-1) = 2k-1$.

This means $n$ has two factors whose factors are of opposite parity.

Observe that powers of $2$ don’t have this property.

$2^0 , 2^1, \cdots 2^6 \in \{1,2, \cdots 100 \}$

The required number is $100 – 7 = 93$

#### IOQM 2021 – Problem 29

Positive integers $a, b, c$ satisfy $\frac{a b}{a-b}=c$. What is the largest possible value of $a+b+c$ not exceeding $99$?

Try out with example.

$b = 18, a =27 , c = 54$

#### IOQM 2021 – Problem 30

Find the number of pairs $(a, b)$ of natural numbers such that $b$ is a $3$ -digit number $a+1$ divides $b-1$ and $b$ divides $a^{2}+a+2$.

$a+1= x$, $b-1=y$

$x|y$

$\Rightarrow kx =y$

$y+1|x^2 -x +2$

$\Rightarrow kx+1|x^2-x+2$

$\Rightarrow kx + 1| kx^2 -kx +2x$

$\Rightarrow kx+1|(kx^2 +x) – (kx +1)-(x -(2x+1))$

$\Rightarrow kx+1|x-(2k+1)$

$\Rightarrow kx+1| kx – k(2k+1)$

$\Rightarrow kx+1|k(2k+1)+1$

$kx+1|k(2k+1)+1$

$b|k(2k+1)+1$

$\Rightarrow b|2k^2+k+1$

$100 \leq b \leq 999$

$\Rightarrow 100 \leq 2k^2 +k+1 \leq 999$

$7 \leq k \leq 22$

By solving this

$16$ possible values.

## Pigeonhole Principle

“The Pigeonhole principle” ~ Students who have never heard may think that it is a joke. Pigeonhole Principle is one of the simplest but most useful ideas in mathematics. Let’s learn the Pigeonhole Principle with some applications.

## Pigeonhole Principle Definition:

In Mathematics, the pigeonhole principle states that if we must put N + 1 or more pigeons into N Pigeon Holes, then some pigeonholes must contain two or more pigeons.

### Pigeonhole Principle Example:

If Kn+ 1 (where k is a positive integer) pigeons are distributed among n holes than some hole contains at least k + 1 pigeons.

### Applications of Pigeonhole Principle:

This principle is applicable in many fields like Number Theory, Probability, Algorithms, Geometry, etc.

## Problems:

### Problem 1

A bag contains beads of two colours: black and white. What is the smallest number of beads which must be drawn from bag, without looking so that among these beads, two are of the same colour?

Solution: We can draw three beads from bags. If there were no more than one bead of each colour among these, then there would be no more than two beads altogether. This is obvious and contradicts the fact that we have chosen there beads. On the other hand, it is clear that choosing two beads is not enough. Here the beads play the role of pigeons, and the colours (black and white) play the role of pigeonhole.

### Problem 2

Find the minimum number of students in a class such that three of them are born in the same month?

Solution: Number of month n =12

According to the given condition,

K+1 = 3

K = 2

M = kn +1 = 2*12 + 1 = 25.

### Problem 3

Show that from any three integers, one can always chose two so that $a^3$b – a$b^3$ is divisible by 10.

Solution: We can factories the term $a^3$b – a$b^3$ = ab(a + b)(a – b), which is always even, irrespective of the pair of integers we choose.

If one of three integers from the above factors is in the form of 5k, which is a multiple of 5, then our result is proved.

If none of the integers are a multiple of 5 then the chosen integers should be in the form of (5k)+-(1) and (5k)+-(2) respectively.

Clearly, two of these three numbers in the above factors from the given expression should lie in one of the above two from, which follows by the virtue of this principle.

These two integers are the ones such that their sum and difference is always divisible by 5. Hence, our result is proved.

### Problem 4

If n is a positive integer not divisible by 2 or 5 then n has a multiple made up of 1’s.

Watch the solution:

Categories

## Indian Olympiad Qualifier in Mathematics – IOQM

Day and Date: Sunday, January 17, 2021

Time of exam: 9 am to 12 noon

Venue: Designated IOQM Centres

Type of exam: Three hour paper and pen exam with responses to be written on OMR sheet. The IOQM will have 30 questions with each question having an integer answer in the range 00-99. The syllabus and standard of this examination will be the same as that of PRMO of previous years. Sample questions can be downloaded from here

Language of the question paper: English and Hindi (Option to be chosen during enrollment)

Due to COVID 19 Pandemic, the Maths Olympiad stages in India has changed. Here is the announcement published by HBCSE:

Important Announcement [Updated:14-Sept-2020]

The national Olympiad programme in mathematics culminating in the International Mathematical Olympiad (IMO) 2021 and European Girls’ Mathematical Olympiad (EGMO) 2022 has been disrupted by the COVID -19 pandemic in the country. In view of the prevailing situation, the following decisions are announced.

❖ In a departure from the usual four-stage procedure for the selection of the teams to represent India at the IMO 2021 and EGMO 2022 that has been followed in previous years, the selection procedure for the 2020-2021 cycle has been condensed to a three-stage process as an exception only for this year. The three stages will be:

1. A three-hour examination called the Indian Olympiad Qualifier in Mathematics (IOQM) organised by the Mathematics Teachers’ Association (India) (MTA(I)).
2. The Indian National Mathematical Olympiad (INMO) organised by the Homi Bhabha Centre for Science Education – Tata Institute of Fundamental Research (HBCSE – TIFR).
3. The International Mathematical Olympiad Training Camp (IMOTC) organised by HBCSE.

❖ The IOQM will have 30 questions with each question having an integer answer in the range 00-99. The syllabus and standard of this examination will be the same as that of PRMO of previous years.

❖ The INMO will be a four-hour examination with 6 questions. The syllabus and standard of this examination will be the same as that of the INMO of previous years.

❖ The IOQM will be conducted by the MTA(I) with support from Indian Association of Physics Teachers (IAPT) and HBCSE.

❖ The INMO and the subsequent stages of the Olympiad programme will be carried out by HBCSE.

❖ Online enrolment for IOQM is expected to start on the IAPT website (iaptexam.in) by October 15, 2020. Further confirmation regarding this and detailed information regarding eligibility and enrolment for IOQM will be announced shortly. These are expected to be similar to previous years.

❖ The tentative schedule for IOQM is Sunday, January 17, 2021 from 9:00 – 12:00 hrs.

❖ The tentative schedule for INMO is Sunday, March 7, 2021 from 12:00 – 16:00 hrs.

❖ Please note that the schedules are tentative, and are subject to change at short notice, depending on the prevailing pandemic situation in the country.

❖ There will not be any examination equivalent to the Regional Mathematical Olympiad for the 2020-2021 cycle. The detailed criteria for qualification from IOQM to INMO and from INMO to IMOTC will be announced soon.

❖ The programme for stages beyond INMO will be announced at an appropriate later time.

HBCSE

Apply for Indian Olympiad Qualifier in Mathematics – IOQM: iaptexam.in

Categories

## Triangle Inequality – Mathematical Circles – Problem No. 5

Try this beautiful problem from Mathematical Circles book based on Triangle inequality.

## Problem :

Find a point inside a convex quadrilateral such that the sum of the distances from the point to the vertices is minimal .

### Key Concepts

Triangle Inequality

Inequality

Geometry

Mathematical Circles – Chapter 6 – Triangle Inequality – Problem 6

Mathematical Circles by Dmitri Fomin , Sergey Genkin , Llia Itenberg

## Try with Hints

Do you really need a hint ? You can start thinking about the Triangle Inequality………..

If you have already get the idea about the main concept for this sum then you can start the problem by taking a quadrilateral ABCD with diagonals that intersect at point ‘o’.

The distance from ‘o’ to all vertices is equal.

Here is the diagram where OA + OB + OC + OD = AC + BD – which is sum of the diagonals. We can consider this as one case …

As a last hint you have to take another point to o’ to compare with first case

Now o’ be another point inside the quadrilateral. If we use triangle inequality here we have ,

AO’ + OC’ > AC from Triangle AO’C

BO’ + O’D > BD from Triangle BO’D

Hence AO’ + O’C + BO’ + O’D > AB + BD

Therefore its clear from this that the point that minimizes the sum of the distances is the point of intersection of diagonals.

Categories

## Cubes and Rectangles – Math Olympiad Hanoi 2018

Try this beautiful problem from Math Olympiad Hanoi, 2018 based on Cubes and Rectangles.

Find the number of rectangles can be formed by the vertices of a cube.

• 6
• 8
• 18
• 12

### Key Concepts

Geometry

Permutation

Combination

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

First hint

There are 6 squares on 6 faces on the cube.

Second Hint

There are 4 diagonals of the cube that have the same length

and pass through the center of the cube. Every two diagonals intersect at the midpoint and form a rectangle.

Final Step

Then there are 6+ $(\frac{4!}{2!2!})$ =12 rectangles.

Categories

## FERMAT POINT

ABC is a Triangle and P be a Fermat Point Inside it.draw three equilateral triangle based on the three sides i.e$\triangle ABA’$, $\triangle ACC’$, $\triangle BCB’$ respectively.Join $AB’$,$BC’$ and$CA’$ .Show that $AB’$,$BC’$ and $CA’$ pass through a single piont i.e they are concurrent.

### Key Concepts

Rotation

Geometry

shortest distance

Challenges and thrills of pre college mathematics

## Try with Hints

Rotation:

ABC is a Triangle . Let P Be any point join $AP,BP$ and $CP$. Now if we rotate the $\triangle ABP$ about the point at B $60 ^{\circ}$ anti clockwise we will get $\triangle BP’A’$.

SHORTEST DISTANCE:

Join the point P and P’.Now In the triangle BPP’ we have

BP-BP’

$\angle PBP’=60 ^{\circ}$, SO $\triangle BPP’$ is a equilateral triangle. so $BP=BP’=PP’$

and also $AP’=AP$ (Length remain unchange after Rotation).

So from the point $A’$ to $C$ the path is $A’P’+PP’+PC$.This path will be Shortest distance if $A’P’+PP’+PC$ i.e A’C be a straight line. and also $AP+PB+AB=A’P’+PP’+PC$

the shortest path betwween two points is a straight line and so $PA+PB+PC$ reaches its minimum if and only if the point $p$ and $P’$ lie on the line $A’C$

By symmetry it follows that $P$ must also lie on the line $BC’$ and $AB’$.

So the point of intersection of these lines is a fermat point of a $\triangle ABC$.

EQUILATERAL TRIANGLE :

Now the triangle $AA’B$ we have

$A’B=AB$ (length remain unchange due to rotation)

$\angle A’BA =60^{\circ}$. so the triangle $AA’B$ is a equilateral triangle .

similarly for the other two triangles $AC’C$ and $BB’C$

Categories

## Can we Prove that ……..

The length of any side of a triangle is not more than half of its perimeter

### Key Concepts

Triangle Inequality

Perimeter

Geometry

Answer: Yes we can definitely prove that by Triangle Inequality

Mathematical Circles – Chapter 6 – Inequalities Problem 3

Mathematical Circles by Dmitri Fomin , Sergey Genkin , Llia Itenberg

## Try with Hints

We can start this sum by using this picture below

The length of the three sides of this triangle are a,b and c. So if we apply triangle inequality which implies that the length of one side of a triangle is less than the sum of the lengths of the two sides of that triangle. In reference to the theorem

b + c > a

So can you try to do the rest of the sum ????????

According to the question we have to find the perimeter at first

Perimeter is the sum of the length of all sides of the triangle = a + b + c

And the length of each side is a or b or c.

We have to prove : a + b + c > length of any one side

This can be one of the most important hint for this problem. Try to do the rest of the sum …………………………..

Here is the rest of the sum :

As stated above if we use triangle inequality :

b + c > a

Lets add a to both the sides

a + b + c > a + a

a + b + c > 2 a

The left hand side of the above inequality is the perimeter of this triangle.

perimeter > 2 a

So , $\frac {perimeter}{2} > a$

$\frac {perimeter}{2}$ = semi perimeter

Hence this is proved that the length of one side of a triangle is less than half of its perimeter.

Categories

## Math Olympiad in India | A Comprehensive Guide

The Math Olympiad Program in India aims to encourage the students interested in Mathematics. It nurtures their talent through healthy competition among pre-university students in India. This programme is one of the major initiatives undertaken by the National Board of Higher Mathematics (NBHM) and is organized by Homi Bhabha Centre for Science Education (HBCSE).

In India, there are 25 regions designated for training and selection of students in these stages. Each of these regions is assigned to a Regional Coordinator (RC). The Indian National Mathematics Olympiad (INMO) leads the Indian Students to participate in the International Math Olympiad (IMO).

## Eligibility for the Math Olympiad

Candidates born on or after August 1, 2001 and studying in Class 8, 9, 10, 11 or 12 are eligible to write PRMO 2020. Further, the candidates must be Indian citizens. Provisionally, students with OCI cards are eligible to write the PRMO, subject to this condition: https://olympiads.hbcse.tifr.res.in/how-to-participate/eligibility/mathematical-olympiad/

## Stages of Math Olympiad in India

Math Olympiad in India is a six-stage process:

• Pre Regional Mathematics Olympiad (Pre RMO): This is the first phase of the International Mathematics Olympiad, conducted in August. In this exam, there are 30 questions that the students need to solve in two and a half hours of examination. Answers are marked on a machine-readable OMR Sheet. Students from Class 8 can enroll.
• Regional Mathematical Olympiad (RMO): This is the second phase of the International Mathematics Olympiad, conducted in October. The question paper contains six problems and the duration of the exam is three hours. The difficulty level of the exam is higher than in the first phase.
• Indian National Mathematical Olympiad (INMO): This is the third phase of the International Mathematics Olympiad, conducted in January. The students selected (approximately 900) in the RMO are eligible for the Indian National Mathematical Olympiad (INMO). It is held at 28 centers across the country. Successful candidates get direct entry in Indian Statistical Institute and Chennai Mathematical Institute interview pool for their prestigious bachelor’s program.
• IMO Training Camp – Approximately best 35 students are invited to take training at HBCSE from April to May. In this camp, stress is laid on building the concepts and problem-solving skills of the students. Also, orientation is provided for the IMO. Students need to take several selection tests during this training period. Taking the performances of the students into consideration, only six students are selected to go for the IMO Examination.
• Pre Departure Camp – The selected students go through rigorous training of 8-10 days before the main Olympiad.
• IMO – A team of 6 candidates accompanied by 4 teachers or mentors represents India in the greatest mathematics contest of the world.

## What is not an actual Math Olympiad?

Many private organizations use the name of IMO to conduct their own contests. These include booksellers who are trying to sell books to coaching centers who use it as a marketing trick. Parents and teachers have requested them to not use the name of IMO.

These private contests are no way similar to the actual math olympiad which is organized by the department of atomic energy and Indian Statistical Institute. In fact, they can sometimes be counterproductive as they promote rote learning of formulas.

In the actual International Math Olympiad, students have 9 hours to solve 6 problems. These problems require special problem-solving skills and creativity. The fake olympiads usually kill this creative spirit and replace them with rote formula learning.

Here is a public petition against fake math olympiads in India.

Taste some Real & Mind-Bending Mathematics Problems with us!

## How to prepare for Math Olympiad?

Math Olympiad has four core topics:

• Number Theory
• Geometry
• Algebra
• Combinatorics

The preparation involves months of problem-solving and concept building. The real math olympian is regarded very highly by the prestigious universities of the world. For example, an IMO medal is almost a sure shot entry to Ivy League universities like Harvard, MIT, Yale. Even an INMO level medal, lets the student take the interview of Indian Statistical Institute’s B.Stat – B.Math Program, and Chennai Mathematical Institute’s B.Sc. Math Program, bypassing the entrance tests.

## Cheenta Program for Maths Olympiad

Cheenta has a rigorous program for Maths Olympiad candidates starting as early as class 1. Our team consists of Olympians and researchers from leading universities in the world.

The unique feature of Cheenta program is that it has both one-on-one and group classes for every student, every week.

Are you in Serious love with Mathematics? Participating in Math Olympiad is a great way to measure your love and who else is a better choice than Cheenta for Math olympiad. Cheenta has an experience of teaching Advanced Mathematics to Outstanding Kids since 10 years from India and abroad.

Don’t believe us? Take the Trial Class for free and decide yourself.

### Miscellaneous

• Challenges and Thrills of Pre-College Mathematics by Venkatchala
• Excursion in Mathematics by Bhaskaracharya Pratishthana
• Problem Solving Strategies by Arthur Engel
• Test of Mathematics at 10+2 Level by East West Press
• IMO Compendium

### Number Theory

• Elementary Number Theory by David Burton
• Elementary Theory of Numbers by W. Sierpinsky

### Combinatorics

• Principles and Techniques in Combinatorics by Chen Chuan Chong and Koh Khee Meng
• Graph Theory by Harary
• Notes by Yufei Zhao

### Algebra

• Polynomials by Barbeau
• Inequality by Little Mathematical Library
• Secrets in Inequalities by Pham Kim Hung
• Complex Numbers from A to Z by Titu Andreescu

### Geometry

• Lines and Curves by Vasiliyev (something else)
• Geometric Transformation by Yaglom
• Notes by Yufei Zhao
• Trigonometric Delights by El Maor
• Trigonometry by S.L. Loney
• 101 Problems in Trigonometry by Titu Andreescu

## Maths Olympiad Curriculum at Cheenta

Number Theory I

This is the first course in elementary number theory:

• NT.I.1 Primes, Divisibility
• NT.I.2 Arithmetic of Remainders
• NT.I.3 Bezout’s Theorem and Euclidean Algorithm
• NT.I.4 Theory of congruence
• NT.I.5 Number Theoretic Functions
• NT.I.6 Theorems of Fermat, Euler, and Wilson
• NT.I.7 Pythagorean Triples
• NT.I.8 Chinese Remainder Theorem

Combinatorics I

This is the first course in combinatorics and elementary counting techniques:

• Com.I.1 Multiplication and Addition rules
• Com.I.2 Bijection Principles
• Com.I.3 Combinatorial Coefficients
• Com.I.4 Inclusion and Exclusion Principles
• Com.I.5 Pigeon Hole Principle
• Com.I.6 Recursions
• Com.I.7 Shortest Route Problems

Algebra I

This is the first course is school algebra. (We assume that the student is familiar with algebraic expressions, and elementary algebraic identities)

• Alg.I.1 Algebraic identities (Sophie Germain, Cube of three etc.)
• Alg.I.2 Mathematical Induction
• Alg.I.3 Binomial Theorem
• Alg.I.4 Linear Equations
• Alg.I.6 Remainder Theorem
• Alg.I.7 Theorems related to roots of an integer polynomial

Geometry I

• Geo.I.1 Locus visualization
• Geo.I.2 Straight Lines
• Geo.I.3 Triangles
• Geo.I.4 Geometric Constructions
• Geo.I.5 Circles

Trigonometry I

• Trig.I.1 Angle and rotation
• Trig.I.2 Half arcs and Half chords – Genesis of trigonometric ratios
• Trig.I.3 Elementary ratios and associated angles
• Trig.I.4 Trigonometric identities
• Trig.I.5 Geometry and trigonometry
• Trig.I.6 Basic properties of Triangles
• Trig.I.7 Compound Angles
• Trig.I.8 Multiple and Submultiple Angles
• Trig.I.9 Trigonometric Series
• Trig.I.10 Height and Distance

Inequality I

This first course in inequality must be preceded by a basic course in algebra.

• Ineq.I.1 Geometric Inequalities
• Ineq.I.2 Arithmetic and Geometric Mean Inequality
• Ineq.I.3 Cauchy Schwarze Inequality
• Ineq.I.4 Titu’s Lemma

Complex Number I

• Complex.I.1 Geometry of Screw Similarity
• Complex.I.2 Field Properties of complex Number
• Complex.I.3 nth roots of unity and Primitive roots
• Complex.I.4 Basic applications to geometry

Intermediate Curriculum

Number Theory II

• NT.II.1 Mobius Inversion Formula
• NT.II.2 Greatest Integer Function
• NT.II.3 Elementary Group Theory
• NT.II.4 Primitive roots and indices
• NT.II.6 Representation of Integers as sum of squares
• NT.II.7 Perfect Numbers

Combinatorics II

• Com.II.1 Chu Shih Chieh’ Identity (Hockey Stick)
• Com.II.2 Multinomial Coeffiecients
• Com.II.3 Advanced Pigeon Holes and Ramsay numbers
• Com.II.4 Catalan Numbers (and advanced bijection)
• Com.II.5 Stirling numbers of second kind
• Com.II.6 Generating functions
• Com.II.7 Non-linear recurrance

Algebra II

• Alg.II.1 Elementary ring and field theory
• Alg.II.2 Eisenstein’s criterion

Geometry II

• Geo.II.1 Barycentric Coordinates
• Geo.II.2 Miquel Point Configuration
• Geo.II.3 Translation
• Geo.II.4 Rotation
• Geo.II.5 Screw Similarity

Inequality II

• Ineq.II.1 Schur’s Inequality
• Ineq.II.2 Rearrangement Inequality
• Ineq.II.3 Jensen’s Inequality
• Ineq.II.4 Bernoulli’s Inequality and Power means

Complex Number II

• Complex.II.1 Cyclotomic Polynomials
• Complex.II.2 Nine Point theorem and other geometric investigations using complex numbers

Number Theory III

• NT.III.1 Thue’s Theorem
• NT.III.2 Square Free Numbers
• NT.III.3 Diophantine Analysis of second and higher degrees
• NT.III.4 Arithmetic Progression whose terms are primes.
• NT.III.5 Trinomial of Euler
• NT.III.6 Scherk and Richart’s Theorem
• NT.III.7 Amicable Numbers
• NT.III.8 Liouville function
• NT.III.9 Roots of polynomials and roots of congruences
• NT.III.10 Numeri Idonai

Combinatorics III

• Com.III.1 Graph Theory
• Com.III.2 Invariance and Extremal Principles
• Com.III.3 Combinatorial Geometry

Algebra III

• Alg.III.1 Polynomials

Geometry III

• Geo.III.1 Inversive Geometry
• Geo.III.2 Advanced Application of complex numbers
• Geo.III.3 Projective Geometry

Inequality III

• Ineq.III.1 Holder and Minkowski’s inequality

Categories

# Understand the problem

Given a circle $\Gamma$, let $P$ be a point in its interior, and let $l$ be a line passing through $P$.  Construct with proof using a ruler and compass, all circles which pass through $P$, are tangent to $\Gamma$, and whose centres lie on $l$.

##### Source of the problem

RMO 2019 Maharashtra and Goa region

Geometry

Easy

# Try these problems first before watching the video or reading the hints:

(Send it to support@cheenta.com. Our priority response is for internal students, however we occasionally try to respond to external students as well). 1. How do you infer that a parallel line needs to be drawn through the center (to the given line AB (L)?  2. Can you find any isosceles triangle in the picture (once one of the little circles is drawn)? 3. How is the second small circle drawn?

# Watch the video

Do you really need a hint? Try it first!

Consider an inversion with respect to a circle with centre $P$. Call this map $f$. Note that, given any point $X$, $f(X)$ is constructible using ruler and compass. Construct the circle $f(\Gamma)$.

Suppose $\Gamma'$ is one of our solutions. Then $f(\Gamma')$ is a line perpendicular to $l=f(l)$ and tangent to $f(\Gamma)$

There can be no more than two lines perpendicular to $l$ and tangent to $f(\Gamma)$. Thus these two lines are the images of our solution circles.

Invert the lines back to get the solution circles.

Your content goes here. Edit or remove this text inline or in the module Content settings. You can also style every aspect of this content in the module Design settings and even apply custom CSS to this text in the module Advanced settings.

# Connected Program at Cheenta

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

# Similar Problems

## IOQM 2021 Problem Solutions

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## Pigeonhole Principle

“The Pigeonhole principle” ~ Students who have never heard may think that it is a joke. Pigeonhole Principle is one of the simplest but most useful ideas in mathematics. Let’s learn the Pigeonhole Principle with some applications. Pigeonhole Principle Definition: In...

## Triangle Problem | PRMO-2018 | Problem No-24

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## Even Parity and Odd Parity

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## Value of Sum | PRMO – 2018 | Question 16

Try this Integer Problem from Number theory from PRMO 2018, Question 16 You may use sequential hints to solve the problem.

## Chessboard Problem | PRMO-2018 | Problem No-26

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## Measure of Angle | PRMO-2018 | Problem No-29

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## Good numbers Problem | PRMO-2018 | Question 22

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## Polynomial Problem | PRMO-2018 | Question 30

Try this Integer Problem from Number theory from PRMO 2018, Question 30 You may use sequential hints to solve the problem.

## Digits Problem | PRMO – 2018 | Question 19

Try this Integer Problem from Number theory from PRMO 2018, Question 19 You may use sequential hints to solve the problem.