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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

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

Try this beautiful Trigonometry Problem based on Triangle from PRMO -2018, Problem 24.

## Triangle Problem – PRMO 2018- Problem 24

If $\mathrm{N}$ is the number of triangles of different shapes (i.e. not similar) whose angles are all integers (in degrees), what is $\mathrm{N} / 100$ ?

,

• $15$
• $22$
• $27$
• $32$
• $37$

Trigonometry

Triangle

Integer

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

Prmo-2018, Problem-24

#### Check the answer here, but try the problem first

$27$

## Try with Hints

#### First Hint

Given that $\mathrm{N}$ is the number of triangles of different shapes. Therefore the different shapes of triangle the angles will be change . at first we have to find out the posssible orders of the angles that the shape of the triangle will be different…

Now can you finish the problem?

#### Second Hint

case 1 : when $x \geq 1$ & $y \geq 3 \geq 1$
$$x+y+z=180$$
$={ }^{179} \mathrm{C}_{2}=15931$
Case 2 : When two angles are same
$$2 x+y=180$$
1,1,178
2,2,176
$\vdots$
89,89,2

#### Solution

But we have one case $60^{\circ}, 60^{\circ}, 60^{\circ}$
$$\text { Total }=89-1=88$$
Such type of triangle $=3(88)$
When 3 angles are same $=1(60,60,60)$
So all distinct angles’s triangles
$$\begin{array}{l} =15931-(3 \times 88)-1 \ \neq 3 ! \ =2611 \end{array}$$
Now, distinct triangle $=2611+88+1$
$=2700 \ N=2700 \ \frac{N}{100}=27 \$

Categories

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

Try this beautiful Problem based on Value of Sum from PRMO 2018, Question 16.

## Value of Sum – PRMO 2018, Question 16

What is the value of $\sum_{1 \leq i<j \leq 10 \atop i+j=\text { odd }}(i-j)-\sum_{1 \leq i<j \leq 10 \atop i+j=\text { even }}(i-j) ?$

• $50$
• $53$
• $55$
• $59$
• $65$

### Key Concepts

Odd-Even

Sum

integer

Answer:$55$

PRMO-2018, Problem 16

Pre College Mathematics

## Try with Hints

We have to find out the sum . Now substitite $i=1,2,3…9$ and observe the all odd-even cases……

Can you now finish the problem ……….

$i=1 \Rightarrow$$1+(2+4+6+8+10-3-5-7-9) =1-4+10=7 i=2 \Rightarrow$$0 \times 2+(3+5+7+9-4-6-8-10)$
$=-4$

$i=3 \Rightarrow$$1 \times 3+(4+6+8+10-5-7-9) =3-3+10=10 i=4 \Rightarrow$$ 0 \times 4+(5+7+9-6-8-10)=-3$
$i=5 \Rightarrow $$1 \times 5+(6+8+10-7-9)=5-2+10 =13 i=6 \Rightarrow$$ 0 \times 6+(7+9-8-10)=-2$
$i=7 \Rightarrow $$1 \times 7+(8+10-9)=7-1+10=16 i=8 \Rightarrow$$ 0 \times 8+(9-10)=-1$
$i=9 \Rightarrow$$1 \times 9+(10)=19 Can you finish the problem…….. Therefore S =(7+10+13+16+19)-(4-3-2-1) =55 ## Subscribe to Cheenta at Youtube Categories ## Chessboard Problem | PRMO-2018 | Problem No-26 Try this beautiful Chessboard Problem based on Chessboard from PRMO – 2018. ## Chessboard Problem – PRMO 2018- Problem 26 What is the number of ways in which one can choose 60 units square from a 11 \times 11 chessboard such that no two chosen square have a side in common? , • $56$ • $58$ • $60$ • $62$ • $64$ ### Key Concepts Game problem Chess board combination ## Suggested Book | Source | Answer #### Suggested Reading Pre College Mathematics #### Source of the problem Prmo-2018, Problem-26 #### Check the answer here, but try the problem first $62$ ## Try with Hints #### First Hint Total no. of squares =121 Out of these, 61 squares can be placed diagonally. From these any 60 can be selected in { }^{61} C_{60} ways =61 Now can you finish the problem? #### Second Hint From the remaining 60 squares 60 can be chosen in any one way Total equal to { }^{61} \mathrm{C}{60}+{ }^{60} \mathrm{C}{60}=61+1=62 ## Subscribe to Cheenta at Youtube Categories ## Measure of Angle | PRMO-2018 | Problem No-29 Try this beautiful Trigonometry Problem based on Measure of Angle from PRMO -2018. ## Measure of Angle – PRMO 2018- Problem 29 Let D be an interior point of the side B C of a triangle ABC. Let l_{1} and l_{2} be the incentres of triangles A B D and A C D respectively. Let A l_{1} and A l_{2} meet B C in E and F respectively. If \angle B l_{1} E=60^{\circ}, what is the measure of \angle C l_{2} F in degrees? , • $25$ • $20$ • $35$ • $30$ • $45$ ### Key Concepts Trigonometry Triangle Angle ## Suggested Book | Source | Answer #### Suggested Reading Pre College Mathematics #### Source of the problem Prmo-2018, Problem-29 #### Check the answer here, but try the problem first $30$ ## Try with Hints #### First Hint According to the questations at first we draw the picture . We have to find out the value of \angle C l_{2} F. Now at first find out $\angle AED$ and $\angle AFD$ which are the exterioe angles of $\triangle BEL_1$ and $\triangle CL_2F$. Now sum of the angles is 180^{\circ} Now can you finish the problem? #### Second Hint \angle E A D+\angle F A D=\angle E A F=\frac{A}{2} \angle A E D=60^{\circ}+\frac{B}{2} \angle A F D=\theta+\frac{C}{2} Therefore \quad \ln \Delta A E F: \frac{A}{2}+60^{\circ}+\frac{B}{2}+\theta+\frac{C}{2}=180^{\circ} 90^{\circ}+60^{\circ}+\theta=180^{\circ} (as sum of the angles of a Triangle is 180^{\circ} Therefore \quad \theta=30^{\circ} ## Subscribe to Cheenta at Youtube Categories ## Good numbers Problem | PRMO-2018 | Question 22 Try this beautiful good numbers problem from Number theory from PRMO 2018, Question 22. ## Good numbers Problem – PRMO 2018, Question 22 A positive integer k is said to be good if there exists a partition of {1,2,3, \ldots, 20} into disjoint proper subsets such that the sum of the numbers in each subset of the partition is k. How many good numbers are there? • 4 • 6 • 8 • 10 • 2 ### Key Concepts Number theorm good numbers subset ## Check the Answer Answer:6 PRMO-2018, Problem 22 Pre College Mathematics ## Try with Hints What is good numbers ? A good number is a number in which every digit is larger than the sum of digits of its right (all less significant bits than it). For example, 732 is a good number, 7>3+2 and 3>2 . Given that k is said to be good if there exists a partition of {1,2,3, \ldots, 20} into disjoint proper subsets such that the sum of the numbers in each subset of the partition is k. Now at first we have to find out sum of these integers {1,2,3, \ldots, 20}. Later create some partitions such that two partitions be disjoint set and sum of the numbers of these partitions be good numbers Can you now finish the problem ………. Sum of numbers equals to \frac{20 \times 21}{2}=210 \& 210=2 \times 3 \times 5 \times 7 So \mathrm{K} can be 21,30,35,47,70,105 Can you finish the problem…….. Case 1 : \mathrm{A}=\{1,2,3,4,5,16,17,18,19,20\}, \mathrm{B}=\{6,7,8,9,10,11,12,13,14,15\} Case 2 : A=\{20,19,18,13\}, B=\{17,16,15,12,10\}, C=\{1,2,3,4,5,6,7,8,9,11,14\} Case 3 : \mathrm{A}=\{20,10,12\}, \mathrm{B}=\{18,11,13\}, \mathrm{C}=\{16,15,9,2\}, \mathrm{D}=\{19,8,7,5,3\}, \mathrm{E}=\{1,4,6,14,17\} Case 4 : A=\{20,10\}, B=\{19,11\}, C=\{18,12\}, D=\{17,13\}, E=\{16,14\}, F=\{1,15,5\}, G=\{2,3,4,6,7,8\} Case 5 : A=\{20,15\}, B=\{19,16\}, C=\{18,17\}, D=\{14,13,8\}, E=\{12,11,10,2\}, F=\{1,3,4,5,6,7,9\} Case 6 : A=\{1,20\}, B=\{2,19\}, C=\{3,18\} \ldots \ldots \ldots \ldots, J=\{10,11\} Therefore Good numbers equal to 6 ## Subscribe to Cheenta at Youtube Categories ## Polynomial Problem | PRMO-2018 | Question 30 Try this beautiful Polynomial Problem from Number theorm from PRMO 2018, Question 30. ## Polynomial Problem – PRMO 2018, Question 30 Let P(x)=a_{0}+a_{1} x+a_{2} x^{2}+\ldots .+a_{n} x^{n} be a polynomial in which a_{i} is non-negative integer for each \mathrm{i} \in{0,1,2,3, \ldots, \mathrm{n}} . If \mathrm{P}(1)=4 and \mathrm{P}(5)=136, what is the value of \mathrm{P}(3) ? • 30 • 34 • 36 • 39 • 42 ### Key Concepts Number theorm Polynomial integer ## Check the Answer Answer:34 PRMO-2018, Problem 30 Pre College Mathematics ## Try with Hints Given that P(x)=a_{0}+a_{1} x+a_{2} x^{2}+\ldots .+a_{n} x^{n} where \mathrm{P}(1)=4 and \mathrm{P}(5)=136. Now we have to find out P(3). Therefore if we put x=1 and x=5 then we will get two relations . Using these relations we can find out a_0 , a_1, a_2 . Can you now finish the problem ………. a_{0}+a_{1}+a_{2}+\ldots \ldots+a_{n}=4 \Rightarrow a_{i} \leq 4 a_{0}+5 a_{1}+5^{2} a_{2}+\ldots+a 5^{n} a_{n}=136 \Rightarrow a_{0}=1+5 \lambda \Rightarrow a_{0}=1 Can you finish the problem…….. Hence 5 a_{1}+5^{2} a_{2}+\ldots \ldots+5^{n} a_{n}=135 a_{1}+5 a_{2}+\ldots 5^{n-1} a_{n-1}=27 \Rightarrow a_{1}=5 \lambda+2 \Rightarrow a_{1}=2 \Rightarrow 5 a_{2}+\ldots .5^{n-1} a_{n-1}=25 a_{2}+5 a_{3}+\ldots .5^{n-2} a_{n-2}=5 \Rightarrow a_{2}=5 \lambda \Rightarrow a_{2}=0 a_{3}+5 a_{4}+\ldots \ldots \ldots+5^{n-3} a_{n-3}=1 a_{3}=1 \Rightarrow a_{4}+5 a_{5}+\ldots .+5^{n-4} a_{n-3}=0 a_{4}=a_{5}=\ldots . a_{n}=0 Hence P(n)=x^{3}+2 x+1 P(3)=34 ## Subscribe to Cheenta at Youtube Categories ## Digits Problem | PRMO – 2018 | Question 19 Try this beautiful Digits Problem from Number theorm from PRMO 2018, Question 19. ## Digits Problem – PRMO 2018, Question 19 Let N=6+66+666+\ldots \ldots+666 \ldots .66, where there are hundred 6 ‘s in the last term in the sum. How many times does the digit 7 occur in the number N ? • 30 • 33 • 36 • 39 • 42 ### Key Concepts Number theorm Digits Problem integer ## Check the Answer Answer:33 PRMO-2018, Problem 19 Pre College Mathematics ## Try with Hints Given that \mathrm{N}=6+66+666+……. \underbrace{6666 …..66}_{100 \text { times }} If you notice then we can see there are so many large terms. but we have to find out the sum of the digits. but since the number of digits are large so we can not calculate so eassily . we have to find out a symmetry or arrange the number so that we can use any formula taht we can calculate so eassily. if we multiply $\frac{6}{9}$ then it becomes =\frac{6}{9}[9+99+\ldots \ldots \ldots \ldots+\underbrace{999 \ldots \ldots \ldots .99}_{100 \text { times }}] Can you now finish the problem ………. \mathrm{N}=\frac{6}{9}[9+99+\ldots \ldots \ldots \ldots+\underbrace{999 \ldots \ldots \ldots .99}_{100 \text { times }}] =\frac{6}{9}\left[(10-1)+\left(10^{2}-1\right)+…….+\left(10^{100}-1\right)\right] =\frac{6}{9}\left[\left(10+10^{2}+…..+10^{100}\right)-100\right] Can you finish the problem…….. =\frac{6}{9}\left[\left(10^{2}+10^{3}+\ldots \ldots \ldots+10^{100}\right)-90\right] =\frac{6}{9}\left(10^{2} \frac{\left(10^{99}-1\right)}{9}\right)-60 =\frac{200}{27}\left(10^{99}-1\right)-60 =\frac{200}{27}\underbrace{(999….99)}_{99 \text{times}}-60 =\frac{1}{3}\underbrace{(222…..200)}_{99 \mathrm{times}}-60 =\underbrace{740740 \ldots \ldots .7400-60}_{740 \text { comes } 33 \text { times }} =\underbrace{740740 \ldots \ldots .740}_{32 \text { times }}+340 \Rightarrow 7 comes 33 times ## Subscribe to Cheenta at Youtube Categories ## Chocolates Problem | PRMO – 2018 | Problem No. – 28 Try this beautiful Problem on Combinatorics from integer based on chocolates from PRMO -2018 ## Chocolates Problem – PRMO 2018- Problem 28 Let N be the number of ways of distributing 8 chocolates of different brands among 3 children such that each child gets at least one chocolate, and no two children get the same number of chocolates. Find the sum of the digits of \mathrm{N}. , • $28$ • $90$ • $24$ • $16$ • $27$ ### Key Concepts Combination Combinatorics Probability ## Suggested Book | Source | Answer #### Suggested Reading Pre College Mathematics #### Source of the problem Prmo-2018, Problem-28 #### Check the answer here, but try the problem first $24$ ## Try with Hints #### First Hint we have to distribute $8$ chocolates among $3$ childrens and the condition is Eight chocolets will be different brands that each child gets at least one chocolate, and no two children get the same number of chocolates. Therefore thr chocolates distributions will be two cases as shown below….. Now can you finish the problem? #### Second Hint case 1:(5,2,1) Out of $8$ chocolates one of the boys can get $5$ chocolates .So $5$ chocolates can be choosen from $8$ chocolates in $8 \choose 5$ ways. Therefore remaining chocolates are $3$ . Out of $3$ chocolates another one of the boys can get $2$ chocolates .So $2$ chocolates can be choosen from $3$ chocolates in $3 \choose 2$ ways. Therefore remaining chocolates are $1$ . Out of $1$ chocolates another one of the boys can get $1$ chocolates .So $1$ chocolates can be choosen from $1$ chocolates in $1 \choose 1$ ways. Therefore number of ways for first case will be $8 \choose 5$ $\times$ $3 \choose 2$ $\times$ $1 \choose 1$$\times$ 3!=\frac{8}{2!.5!.1!}$$\times 3$

Case 2:$(4,3,1)$

Out of $8$ chocolates one of the boys can get $4$ chocolates .So $4$ chocolates can be choosen from $8$ chocolates in $8 \choose 4$ ways.

Therefore remaining chocolates are $4$ . Out of $4$ chocolates another one of the boys can get $3$ chocolates .So $3$ chocolates can be choosen from $4$ chocolates in $4 \choose 3$ ways.

Therefore remaining chocolates are $1$ . Out of $1$ chocolates another one of the boys can get $1$ chocolates .So $1$ chocolates can be choosen from $1$ chocolates in $1 \choose 1$ ways.