IOQM 2024 - Problems and Solutions

IOQM is the first level of (real) math olympiad in India. More than a 100,000 students take it every year. Success in IOQM leads to the second level which is RMO (Regional Math Olympiad) and third level INMO (Indian National Math Olympiad).

In this post we have added the problems and solutions from the IOQM 2024.

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Problem 1

The smallest positive integer that does not divide $ \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9$ is:

Solution

Since this is 9 ! it will be divisible by all the numbers from $1-9$. Also, 9 ! contains $2 * 5=10$, it is divisible by 10 . Lets check for 11 . Since 11 is a prime number : 9! will not be divisible by 11 .
So, the smallest positive number that does not divides $1 \times 2 \times \times 3 \times 4 \times 5 \times 6 \times$ $7 \times 8 \times 9$ is 11 .

Problem 2

The number of four-digit odd numbers having digits $1,2,3,4$, each occuring exactly once, is:

Solution

Number of odd numbers that we can form using $1,2,3,4$ without repeating implies that ones digit is either 1 or 3 .
Rest of the three places can be occupied any of the remaining three digits. So the number of total ways $=3 \times 2 \times 1 \times 2=12$.

Problem 3

The number obtained by taking the last two digits of $5^{2024}$ in the same order is:

Solution

Last two digits of $5^n=25$

Problem 4

Let $A B C D$ be a quadriateral with $\angle A D C=70^{\circ}, \angle A C D=70^{\circ}, \angle A C B=10^{\circ}$ and $\angle B A D=110^{\circ}$. The measure of $\angle C A B$ (in degrees) is:

Problem 5

Let $a=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$, let $b=\frac{x}{z}+\frac{y}{x}+\frac{z}{y}$ and let $c=\left(\frac{x}{y}+\frac{y}{z}\right)\left(\frac{y}{z}+\frac{z}{x}\right)\left(\frac{z}{x}+\frac{x}{y}\right)$. The value of $|a b-c|$ is:

Solution

$a=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=\frac{x^2 z+y^2 x+z^2 y}{x y z}$

$b=\frac{x}{z}+\frac{y}{x}+\frac{z}{y}=\frac{x^2 y+y^2 z+z^2 x}{x y z}$

$a b=\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)\left(\frac{x}{z}+\frac{y}{x}+\frac{z}{y}\right)$

$=\frac{y^2}{x z}+\frac{x y}{z^2}+1+1 \frac{y^2}{z x}+\frac{z y}{x^2}+\frac{x z}{y^2}+1+\frac{z^2}{x y}$

$c=\left(\frac{x}{y}+\frac{y}{z}\right)\left(\frac{y}{z}+\frac{z}{x}\right)\left(\frac{z}{x}+\frac{x}{y}\right)$

$=1+\frac{y^2}{z x}+ \frac{z^2}{x y}+\frac{y z}{x^2}+\frac{x^2}{z x}+\frac{x y}{z^2}+\frac{x z}{y^2}+1$

=$\left(\frac{x}{z}+\frac{y^2}{z^2}+\frac{z}{y}+\frac{y}{x}\right)\left(\frac{z}{x}+\frac{x}{y}\right)$

$=1+\frac{y^2}{z x}+\frac{z^2}{x y}+\frac{y z}{x^2}+\frac{x^2}{z x}+\frac{x y}{z^2}+\frac{x z}{y^2}+1.$

$|a b-c|=\left\lvert\,\left(\frac{y^2}{x z}+\frac{x y}{z^2}+1+1 \frac{y^2}{z x}+\frac{z y}{x^2}+\frac{x z}{y^2}+1+\frac{z^2}{x y}\right)-\right.$

$1+\frac{y^2}{z x}+\frac{z^2}{x y}+\frac{y z}{x^2}+\frac{x^2}{z x}+\frac{x y}{z^2}+\frac{x z}{y^2}+1$|

$=1 $

Problem 6

Find the number of triples of real numbers $(a, b, c)$ such that $a^{20}+b^{20}+c^{20}=a^{24}+b^{24}+c^{24}=1$

Solution

The triples $(a, b, c)$ is $(1,0,0),(0,1,0),(0,0,1),(-1,0,0),(0,-1,0),(0,0,-1)$

Problem 7

Determine the sum of all possible surface areas of a cube two of whose vertices are $(1,2,0)$ and $(3,3,2)$.

Solution

Notice that the distance between the points $(1,2,0)$ and $(3,3,2)=$ $\sqrt{2^2+1^2+2^2}=3$.

In the cube, we can have the vertices $(1,2,0)$ and $(3,3,2)$ as endpoints of a side of a cube (or) endpoints of a face diagonal (or) endpoints of a main diagonal and there is no other configuration. If $a$ denotes the side length of a cube, then $a=3, a \sqrt{2}=3, a \sqrt{3}=3$ are the only possibilities.

This implies $a=3, \frac{3}{\sqrt{2}}, \sqrt{3}$. We know that the surface area of cube $=$ $6 a^2=6 \times 9$ (or) $6 \times \frac{9}{2}$ (or) $6 \times 3$. Hence, the sum of all possible surface areas $=6 \times 9+6 \times \frac{9}{2}+6 \times 3=6\left(9+\frac{9}{2}+3\right)=99$.

Problem 8

Let $n$ be the smallest integer such that the sum of digits of $n$ is divisible by 5 as well as the sum of digits of $(n+1)$ is divisible by 5 . What are the first two digits of $n$ in the same order?

Solution

Let $S(m)$ denote the sum of digits of a natural number $m$. Notice that, $S(m+1)=S(m)+1$, if $m$ does not end with 9 .

Clearly, if the number $n$ does not end with 9 , then $S(n)$ and $S(n+1)$ would only differ by 1 and hence both of them cannot be divisible by 5 simultaneously. Hence, $n=\overline{x 9}$.

Now, similarly if $x$ does not end with 9 , then $S(n)=S(x)+9 ; S(n+1)=$ $S(x+1)=S(x)+1 \Rightarrow S(n+1)-S(n)=8$, contradiction. Hence, $n=\overline{y 99}$. And again $y$ must end with 9 as otherwise, $S(n+1)-S(n)=9+9-1=17$, contradiction.

If we proceed in this way, then we would get $n=\overline{z 9999}$. Now, for the smallest value of such $n$, we take $z$ to be a single digit number. Note that $9+9+9+9=36$ is 1 modulo 5 and hence we need $z$ to be 4 modulo 5 . Thus, $z=4$ is the least possible such number and hence $n=49999$ is the least possible value of $n$.

Problem 9

Consider the grid of points $X={(m, n) \mid 0 \leq m, n \leq 4}$. We say a pair of points ${(a, b),(c, d)}$ in $X$ is a knight-move pair if $(c=a \pm 2$ and $d=b \pm 1$ ) or ( $c=a \pm 1$ and $d=b \pm 2$ ). The number of knight-move pairs in $X$ is:

Solution

Solution: We will find the possible number of knight-move pairs corresponding to each point and add all of them. But in this counting each point would be over counted exactly twice (as pair consits of two distinct points) and we divide by 2 at last in order to account for it and to get the final answer.

Let us consider only the points in $Y={(m, n) \mid 0 \leq m, n \leq 2} \subset X$ because using symmetry one can extend the number of possible pairs containing that point to all the other points. Counting by this way explicitly, one would get the number of pairs as shown in the diagram.

Hence, the sum of all the numbers $=4 \times(2+3+4+3)+4 \times(4+6)+8=96$. So, final answer $=\frac{96}{2}=48$.

Problem 10

Determine the number of positive integral values of $p$ for which there exists a triangle with sides $a, b$, and $c$ which satisfy

$$
a^2+\left(p^2+9\right) b^2+9 c^2-6 a b-6 p b c=0 .
$$

Solution

Notice that

$$
a^2+\left(p^2+9\right) b^2+9 c^2-6 a b-6 p b c=(p b-3 c)^2+(a-3 b)^2=0
$$

Hence, individually the bracket values must be $0 \Rightarrow p b-3 c=0 ; a-3 b=0 \Rightarrow$ $a=3 b ; p b=3 c$. Hence, $(a, b, c)=\left(3 b, b, \frac{p b}{3}\right)$. We know that the necessary and sufficient condition for existence of a triangle is that the largest side being less than sum of other two sides. So, we take two cases now,

If $p<9 \Rightarrow \frac{p b}{3}<3 b \Rightarrow 3 b$ is the largest side. Hence, we need $3b<b+\frac{pb}{3}\Rightarrow p>6$. Hence, the allowed values for this case are $p=7,8$.

If $p \geq 9 \Rightarrow \frac{p b}{3}\geq 3 b \Rightarrow \frac{p b}{3}$ is the largest side. Hence, we need $\frac{p b}{3}<3 b+b \Rightarrow$ $p<12$. Hence, the possible values for this case are $p=9,10,11$.

Hence, there five possible values for $p$ which are $7,8,9,10,11$.

Problem 11

The positive real numbers $a, b, c$ satisfy:

$$
\begin{aligned}
& \frac{a}{2 b+1}+\frac{2 b}{3 c+1}+\frac{3 c}{a+1}=1 \
& \frac{1}{a+1}+\frac{1}{2 b+1}+\frac{1}{3 c+1}=2
\end{aligned}
$$

What is the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} ?$

Solution

By adding up the two equations, we would get,
$$\frac{a+1}{2b+1}+\frac{2b+1}{3c+1}+\frac{3c+1}{a+1}=3\rightarrow Eq 1$$
Notice that each of these terms are positive real numbers and their product $\frac{a+1}{2b+1}\times\frac{2b+1}{3c+1}\times\frac{3c+1}{a+1}=1$.

Let $\alpha=\frac{a+1}{2b+1},\ \beta=\frac{2b+1}{3c+1},\ \gamma=\frac{3c+1}{a+1}$, then by the AM-GM inequality, $$\frac{\alpha+\beta+\gamma}{3}\geq\left(\alpha\beta\gamma\right)^{\frac{1}{3}}=1$$
with the equality holding true when $\alpha=\beta=\gamma=1$. But from the equation $Eq 1$, we know that the equality holds true and hence $\frac{a+1}{2b+1}=\frac{2b+1}{3c+1}=\frac{3c+1}{a+1}=1\Rightarrow \boxed{a=2b=3c}$. Now, substituting it in the first equation in the question, we get, $a=\frac{1}{2}=2b=3c\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=12$.

Problem 12

Consider a square $A B C D$ of side length 16. Let $E, F$ be points on $C D$ such that $C E=E F=F D$. Let the line $B F$ and $A E$ meet in $M$. The area of $\triangle M A B$ is:

Solution

Drop perpendiculars from $M$ to $AB, CD$ and let the foot of perpendiculars be $X,Y$ respectively as shown.

Let $a=16=side\ of\ square$. Note that $CE=EF=FD\Rightarrow EF=\frac{a}{3}$. Since, $AB||EF\Rightarrow\triangle ABM\sim\triangle EFM\Rightarrow\frac{MX}{MY}=\frac{AB}{EF}=3$. But, we know that $XY=a\Rightarrow MX=\frac{3a}{4},\ MY=\frac{a}{4}$. Hence, area of $\triangle MAB=\frac{1}{2}\times MX\times AB=\frac{3a^2}{8}=96$.

Problem 13

Three positive integers $a, b, c$ with $a>c$ satisfy the folowing equations:

$$
a c+b+c=b c+a+66, \quad a+b+c=32
$$

Find the value of $a$.

Solution

Problem 14

Initially, there are $3^{80}$ particles at the origin $(0,0)$. At each step the particles are moved to points above the $x$-axis as follows: if there are $n$ particles at any point $(x, y)$, then $\left\lfloor\frac{n}{3}\right\rfloor$ of them are moved to $(x+1, y+1),\left\lfloor\frac{n}{3}\right\rfloor$ are moved to $(x, y+1)$ and the remaining to $(x-1, y+1)$. For example, after the first step, there are $3^{79}$ particles each at $(1,1),(0,1)$ and $(-1,1)$. After the second step, there are $3^{78}$ particles each at $(-2,2)$ and $(2,2), 2 \times 3^{78}$ particles each at $(-1,2)$ and $(1,2)$, and $3^{79}$ particles at $(0,2)$. After 80 steps, the number of particles at $(79,80)$ is:

Solution

Problem 15

Let $X$ be the set consisting of twenty positive integers $n, n+2, \ldots, n+38$. The smallest value of $n$ for which any three numbers $a, b, c \in X$, not necessarily distinct, form the sides of an acute-angled triangle is:

Problem 16

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function satisfying the relation $4 f(3-x)+3 f(x)=x^2$ for any real $x$. Find the value of $f(27)-f(25)$ to the nearest integer. (Here $\mathbb{R}$ denotes the set of real numbers.)

Solution

Problem 17

Consider an isosceles triangle $A B C$ with sides $B C=30, C A=A B=20$. Let $D$ be the foot of the perpendicular from $A$ to $B C$, and let $M$ be the midpoint of $A D$. Let $P Q$ be a chord of the circumcircle of triangle $A B C$, such that $M$ lies on $P Q$ and $P Q$ is parallel to $B C$. The length of $P Q$ is:

Solution

Problem 18

Let $p, q$ be two-digit numbers neither of which are divisible by 10 . Let $r$ be the four-digit number by putting the digits of $p$ followed by the digits of $q$ (in order). As $p, q$ vary, a computer prints $r$ on the screen if $\operatorname{gcd}(p, q)=1$ and $p+q$ divides $r$. Suppose that the largest number that is printed by the computer is $N$. Determine the number formed by the last two digits of $N$ (in the same order).

Problem 19

Consider five points in the plane, with no three of them collinear. Every pair of points among them is joined by a line. In how many ways can we color these lines by red or blue, so that no three of the points form a triangle with lines of the same color.

Solution

Problem 20

On a natural number $n$ you are allowed two operations: (1) multiply $n$ by 2 or (2) subtract 3 from n. For example starting with 8 you can reach 13 as follows: $8 \rightarrow 16 \rightarrow 13$. You need two steps and you cannot do in less than two steps. Starting from 11, what is the least number of steps required to reach 121 ?

Solution

Problem 21

An intenger $n$ is such that $\left\lfloor\frac{n}{9}\right\rfloor$ is a three digit number with equal digits, and $\left\lfloor\frac{n-172}{4}\right\rfloor$ is a 4 digit number with the digits $2,0,2,4$ in some order. What is the remainder when $n$ is divided by $100 ?$

Solution

Problem 22

In a triangle $A B C, \angle B A C=90^{\circ}$. Let $D$ be the point on $B C$ such that $A B+B D=A C+C D$. Suppose $B D: D C=2: 1$. If $\frac{A C}{A B}=\frac{m+\sqrt{p}}{n}$, where $m, n$ are relatively prime positive integers and $p$ is a prime number, determine the value of $m+n+p$.

Solution

Problem 23

Consider the fourteen numbers, $1^4, 2^4, \ldots, 14^4$. The smallest natural number $n$ such that they leave distinct remainders when divided by $n$ is:

Solution

Observe that, $14^4 \equiv(-1)^4 \equiv 1^4(\bmod 15)$

similarly, $14^4 \equiv(-2)^4 \equiv 2^4(\bmod 16)$

...

$14^4 \equiv(-13)^4 \equiv 13^4(\bmod 27)$

$\therefore n \geq 28$

for $\mathrm{n}=28: 13^4 \equiv 1^4(\bmod 28)$
for $\mathrm{n}=29$ and 30 we have

$5^4 \equiv 2^4(\bmod 29)$

$4^4 \equiv 2^4(\bmod 30)$ respectively

Now if $\mathrm{a}^4 \equiv \mathrm{b}^4(\bmod 31)$

$\Rightarrow \mathrm{a}^{32} \equiv \mathrm{b}^{32}(\bmod 32)$

using Fermat’s Little theorem we get $a \equiv b(\bmod 31)$ which is not possible hence
31 is the right answer

Problem 24

Consider the set $F$ of all polynomials whose coefficients are in the set of ${0,1}$. Let $q(x)=x^3+x+1$. The number of polynomials $p(x)$ in $F$ of degree 14 such that the product $p(x) q(x)$ is also in $F$ is:

Solution

Problem 25

A finite set $M$ of positive integers consists of distinct perfect squares and the number 92 . The average of the numbers in $M$ is 85 . If we remove 92 from $M$, the average drops to 84 . If $N^2$ is the largest possible square in $M$, what is the value of $N$ ?

Solution

Let $\mathrm{M}={a_1^2, a_2^2 \ldots \ldots . a_n^2, 92}$
$\Rightarrow \frac{\sum a_1^2}{n}=84, \frac{\sum a_1^2+9^2}{n+1}=85$
$\Rightarrow \frac{84 n+9^2}{n+1}=85 \quad \Rightarrow n=7 \text {. }$
$\text { Note that } 1^2+2^2+\ldots .+6^2=91$
$\sum a_i^2=588 \quad \Rightarrow N \leq 22 \text { but } 1^2+2^2+3^2+4^2+5^2+7^2+22^2=588$
$\Rightarrow N=22$

Solution

Let $\mathrm{M}={a_1^2, a_2^2 \ldots \ldots \ldots a_n^2, 92}$

$\frac{\sum a_1^2}{n}=84, \frac{\sum a_1^2+9^2}{n+1}=85$

$\frac{84 n+9^2}{n+1}=85 \quad \Rightarrow n=7$

Note that $1^2+2^2+\ldots \ldots+6^2=91$

$\sum a_i^2=588 \quad \Rightarrow N \leq 22$

but $1^2+2^2+3^2+4^2+5^2+7^2+22^2=588$

$\Rightarrow N=22$

Problem 26

The sum of $\lfloor x\rfloor$ for all real numbers $x$ satisfying the equation $16+15 x+15 x^2=\lfloor x\rfloor^3$ is:

Solution

$16+15 x+15 x^2=[x]^9 \leq x^3 $
$\Rightarrow x \geq 16$

$\text { Also, } 16+15 x+15 x^2=[x]^3>(x-1)^3 \Rightarrow x<19$

$x=16$ satisfies the equation
if $\left.\lfloor x\rfloor=17 \Rightarrow 16+15 x+15 x^2\right)=17^3$ has a root in $(17,18)$
But $\lfloor x\rfloor=18$ has no solution
$\text { sum }=16+17=33$

Problem 27

In a triangle $A B C$, a point $P$ in the interior of $A B C$ is such that $\angle B P C-\angle B A C=\angle C P A-\angle C B A=\angle A P B-\angle A C B$. Suppose $\angle B A C=30^{\circ}$ and $A P=12$. Let $D, E, F$ be the feet of perpendiculars form $P$ on to $B C, C A, A B$ respectively. If $m \sqrt{n}$ is the area of the triangle $D E F$ where $m, n$ are integers with prime, then what is the value of the product $m n$ ?

Solution

let $\alpha, \beta, \gamma$ be $\angle B P C, \angle C P A, \angle A P C$ respectively.

$\alpha-A =\beta-B=\gamma-C=\lambda$

$ \Rightarrow \alpha+A+\gamma-(A+B+C)=3 \lambda$

$\Rightarrow \lambda=60 $

$ \Rightarrow \angle B P C=90 $

$A P=12 $

$\Rightarrow E F=12 \sin 30=6 .$

Height of $\Delta=3 \sqrt{3}$

$\text { Area }=\frac{1}{2} \times 6 \times 3 \sqrt{3}=9 \sqrt{3} $
$m n=27$

Problem 28

Find the largest positive integer $n<30$ such that $\frac{1}{2}\left(n^8+3 n^4-4\right)$ is not divisible by the square of ann prime number.

Solution

$f(n)=\frac{n^8+3 n^4-4}{2}$

Note, by Fermat’s little theorem,

$ x^4=1(\bmod 5) $

$ x^8=1(\bmod 5)$

Also, powers of $n$ are even. $n=k(\bmod 25) \quad$ when $k=1,2,3,4$

then $25 / f(n) \Rightarrow n \neq 21,22,23,24,26,27,28,29$
Also note $4 / f(25)$
$\therefore$ largest such $n=20$

Problem 29

Let $n=2^{19} 3^{12}$. Let $M$ denote the number of positive divisors of $n^2$ which are less than $n$ but would not divide $n$. What is the number formed by taking the last two digits of $M$ (in the same order)?

Solution

$n=2^{19} \cdot 3^{12}$

$n^2=2^{19} \cdot 3^{24}$ Let $2^a 3^b$ be such a divisor of $n^2$ which does not divide $n$ but is less than $n$.

Case 1. $a>19, b \leq 12=118$
Case 2. $a \leq 19, b>12=110$

$\therefore 118+110=228 \rightarrow 28$

Problem 30

Let $A B C$ be a right-angled triangle with $\angle B=90^{\circ}$. Let the length of the altitude $B D$ be equal to 12 What is the minimum possible length of $A C$, given that $A C$ and the perimeter of triangle $A B C$ are integers?

Solution

Let

$ C D=x, A D=y$

$ \Rightarrow x+y \geq 2 \sqrt{x y} \quad(\text { AM-GM) }$

$ \Rightarrow x+y \geq 24.$

There are no Pythagorean triples with hypotenuse = 24.

$\therefore$ least possible $A C=25$

(15, 20, 25 triangle)