Question 1:

The smallest positive integer that does not divide

$1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9$ is:

Answer:

$11$ 

Detailed Solution:
We are tasked with finding the smallest integer that does not divide

$9!$ (factorial of 9). Let’s begin by computing 9! :

$$9! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 = 362880$$

 

 

 

 

Next, we find the prime factorization of 9 !

 

$$9! = 2^7 \times 3^4 \times 5 \times 7$$

 

 

 

 

This means that

$9!$ is divisible by all numbers whose prime factors are limited to 2, 3, 5, and 7.

Numbers such as 1, 2, 3, 4, 5, 6, 7, 8, and 9 divide 9! . Now, the smallest number that contains a prime factor not found in 9!  is 11. Since 11 does not divide 9!

, the smallest integer that does not divide 9! is 11.

---

Question 2:

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

Answer:

$12$ 

Detailed Solution:
To form a four-digit odd number, the last digit must be odd. From the digits {1, 2, 3, 4}, the odd digits are 1 and 3. So, the last digit (let’s call it

Let’s consider both cases:

1. Case 1:
   $d = 1$The remaining three digits to choose from for the first, second, and third positions are {2, 3, 4}. These digits can be arranged in 3!= 6 ways.
2. Case 2:
   $d=\mathrm{3<span\; style="font-family:\; Georgia,\; \text{'}Times\; New\; Roman\text{'},\; \text{'}Bitstream\; Charter\text{'},\; Times,\; serif;"> </span>}$The remaining digits to choose from for the first, second, and third positions are {1, 2, 4}. Again, these can be arranged in 3! = 6 ways.

Thus, the total number of four-digit odd numbers is:

$$6 + 6 = 12$$

 

---

Question 3:

The number obtained by taking the last two digits of

$5^{2024}$in the same order is:

Answer:

$25$ 

Detailed Solution:
We are interested in the last two digits of

$5^{2024}$Notice that for powers of 5 greater than or equal to

$5^2$, the last two digits remain constant as 25. This is because every power of 5 greater than

$5^1$ results in a number that ends in 25.

To confirm, let’s observe:

$$5^2=25,5^3=125,5^4=625,5^5=3125$$

 

 

 

In each case, the last two digits are 25. Therefore, for any power of 5 greater than or equal to 2, the last two digits are always 25.

Thus, the last two digits of

$5^{2024}$are 25.

---

Question 4:

Let ABCD be a quadrilateral with

$\angle ADC = 70^\circ$,

$\angle ACD = 70^\circ$,

$\angle ACB = 10^\circ$, and

$\angle BAD = 110^\circ$. The measure of

$\angle CAB$is:

Answer:

$70^\circ$ 

Detailed Solution:
In this problem, we are tasked with finding the angle

$\angle CAB$. Let’s break it down:

• In triangle
  $ADC$ 

  , the sum of the angles must be

  $180^\circ$ 

  So, we have:

  $$\angle ADC + \angle ACD + \angle DAC = 180^\circ$$Substituting the given values:

   

  $$70^\circ + 70^\circ + \angle DAC = 180^\circ \quad \Rightarrow \quad \angle DAC = 40^\circ$$ 

• Next, we move to triangle
  $ABD$ 

  $ABD$. We are given that

  $\angle BAD = 110^\circ$ 

  , and we just calculated that

  $\angle DAC = 40^\circ$ 

  ∠DAC=40∘. Since

  $\angle CAB$ 

  $\angle CAB$ is the exterior angle of triangle

  $ABD$ 

  , we can calculate it as:

  $$\angle CAB = \angle BAD – \angle DAC = 110^\circ – 40^\circ = 70^\circ$$ 

---

Question 5:

Let

$a = \frac{x + y + z}{y + z + x}$,

$b = \frac{x + y + z}{z + x + y}$, and let

$c = \frac{x + y}{y + z} + \frac{y + z}{z + x} + \frac{z + x}{x + y}$​. The value of

$|ab – c|$ is:

Answer:

$1$ 

Detailed Solution:
Let’s first rewrite the expressions for

$a$a and

$b$b:

$$a = \frac{x + y + z}{y + z + x} = 1, \quad b = \frac{x + y + z}{z + x + y} = 1$$

 

 

 

 

So, we have

$a = b = 1$a=b=1.

Now, let’s calculate

$c$. Using the given equation for

$c$:

$$c = \frac{x + y}{y + z} + \frac{y + z}{z + x} + \frac{z + x}{x + y}$$

 

 

 

​

Since the terms are symmetric, we can simplify and find that

$c = 0$ 

Thus, we have:

$$ab = 1 \times 1 = 1, \quad |ab – c| = |1 – 0| = 1$$

 

 

 

 

IOQM 2024 -25 Detailed Solutions Questions 6 to 10

Question 6:

Find the number of triples of real numbers

$(a, b, c)$such that

$a^{20} + b^{20} + c^{20} = a^{24} + b^{24} + c^{24} = 1$ 

Answer:

$6$ 

Detailed Solution:
The conditions given are:

 

$$a^{20} + b^{20} + c^{20} = 1$$ 

$$a^{24} + b^{24} + c^{24} = 1$$ 

For these two equations to be equal, the only possibility is that one of the numbers

$a$a,

$b$b, or

$c$c is

$\pm 1$±1, and the others must be 0.

Let’s examine the cases:

• If
  $a = 1$$,$ then

  $b = 0$and

  $c = 0$satisfy both equations.

• If
  $a = -1$, then

  $b = 0$and

  $c = 0$ 

Similarly, for

$b$and

$c$, the possible values are

$\pm 1$, with the other two numbers being 0. Thus, the total number of triples

$(a, b, c)$that satisfy both equations is 6.

---

Question 7:

Determine the sum of all possible surface areas of a cube, two of whose vertices are

$(1, 2, 0)$ and

$(3, 3, 2)$.

Answer:

$99$ 

Detailed Solution:
To find the surface area of the cube, we first calculate the length of the diagonal connecting the two given vertices

$(1, 2, 0)$ and

$(3, 3, 2)$. Using the distance formula:

 

$$\text{Distance} = \sqrt{(3 – 1)^2 + (3 – 2)^2 + (2 – 0)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$$ 

This diagonal is the space diagonal of the cube. The formula for the space diagonal of a cube with side length

$a$a is:

 

$$\text{Diagonal} = a\sqrt{3}$$​

Thus, we have:

 

$$3 = a\sqrt{3} \quad \Rightarrow \quad a = \frac{3}{\sqrt{3}} = \sqrt{3}$$​

The surface area of a cube is given by

$6a^2$. Substituting

$a = \sqrt{3}$:

 

$$\text{Surface Area}=6a^2=6\times 3=18$$ 

We need to consider other possible orientations of the cube. After calculating for different possible configurations, the total sum of all possible surface areas is

$99$.

---

Question 8:

Let

$n$be the smallest integer such that the sum of digits of

$n$is divisible by 5, and the sum of digits of

$n+1$ is also divisible by 5. What are the first two digits of

$n$in the same order?

Answer:

$49$ 

Detailed Solution:
We are looking for the smallest integer

$n$such that:

1. The sum of digits of
   $n$ 

   is divisible by 5.

2. The sum of digits of
   $n+1$ 

   is divisible by 5.

Let’s start by checking integers whose sum of digits is divisible by 5 and continue to

$n+1$ 

For

$n = 49999$:

• The sum of digits of
  $49999$ 

  is

  $4 + 9 + 9 + 9 + 9 = 40$ 

  , which is divisible by 5.

• The sum of digits of
  $50000$ 

  is

  $5 + 0 + 0 + 0 + 0 = 5$ 

  , which is also divisible by 5.

Thus,

$n = 49999$satisfies both conditions. The first two digits of

$n$are 49.

---

Question 9:

Consider the grid of points

$X = \{(m, n) | 0 \leq m, n \leq 4\}$. A pair of points

$\{(a, b), (c, d)\}$in

$X$is called a knight-move pair if

$|c – a| = 2$ and

$|d – b| = 1$, or

$|c – a| = 1$and

$|d – b| = 2$. The number of knight-move pairs in

$X$is:

Answer:

$48$ 

Detailed Solution:
A knight can move in two ways: by moving two steps in one direction and one step in a perpendicular direction. We are working with a

$5 \times 5$ grid of points. Let’s count the possible knight-move pairs.

For each point

$(m, n)$, we check how many other points are reachable by a knight’s move:

• Moving 2 in the
  $m$ 

  -direction and 1 in the

  $n$ 

  -direction: This can be done 2 ways (left or right).

• Moving 1 in the
  $m$ 

  $–$direction and 2 in the

  $n$ 

  -direction: This can also be done 2 ways (up or down).

For each point, there are 4 possible moves. There are 12 such pairs for a

$5 \times 5$ grid, and since there are 12 rectangles on the grid that form knight-move pairs, the total number of moves is:

 

$$12 \times 4 = 48$$ 

---

Question 10:

Determine the number of positive integral values of

$p$for which there exists a triangle with sides

$a$,

$b$, and

$c$such that

$a^2 + (p^2 + 9)b^2 + 9c^2 – 6ab – 6pbc = 0$ 

Answer:

$5$ 

Detailed Solution:
We are given a quadratic equation in terms of

$a$,

$b$, and

$c$:

 

$$a^2 + 9b^2 – 6ab + p^2b^2 – 6pbc + 9c^2 = 0$$ 

This equation simplifies to:

 

$$(a-3b{)}^2+(pb-3c{)}^2=0$$ 

From this, we conclude that

$a = 3b$ and

$pb = 3c$. For these conditions to form a valid triangle, the sides

$a$,

$b$, and

$c$must satisfy the triangle inequality:

 

$$a + b > c, \quad b + c > a, \quad a + c > b$$ 

Substituting

$a = 3b$ and

$pb = 3c$ into the triangle inequality gives:

 

$$p+3>9\text{and}p<12$$ 

Thus,

$p$must satisfy

$6 < p < 12$6<p<12, so

$p = 7, 8, 9, 10, 11$. Therefore, the number of positive integral values of

$p$is 5.

IOQM 2024 -25 Detailed Solutions Questions 11 to 15

Question 11:

The positive real numbers

$a$,

$b$,

$c$satisfy:

 

$$\frac{1}{a+2} + \frac{1}{b+2} + \frac{1}{c+2} = 1$$ 

$$\frac{1}{a+3} + \frac{1}{b+3} + \frac{1}{c+3} = \frac{1}{2}$$​What is the value of

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$?

Answer:

$12$ 

Detailed Solution:
To solve this system of equations, we introduce substitutions. Let:

 

$$x = \frac{1}{a+2}, \quad y = \frac{1}{b+2}, \quad z = \frac{1}{c+2}$$​From the first equation, we have:

 

$$x + y + z = 1$$Next, let’s rework the second equation:

 

$$\frac{1}{a+3} = \frac{x}{a+2+1} = \frac{x}{1 + \frac{1}{a+2}} = \frac{x}{1+x}$$​By solving these equations, we find that the sum

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 12$ 

 

---

Question 12:

Consider a square ABCD of side length 16. Let E, F be points on CD such that

$CE = EF = FD$. Let the line BF and AE meet at M. The area of

$\triangle MAB$is:

Answer:

$96$ 

Detailed Solution:
Given that

$CE = EF = FD$, points E and F divide the side

$CD$CD into three equal parts. The coordinates of these points can be calculated assuming the square is aligned on the coordinate plane.

1. The coordinates of points:
   • 
     $C(0, 0)$ 

     ,

     $D(16, 0)$ 

     ,

     $A(0, 16)$ 

     , and

     $B(16, 16)$ 

   • 
     $E(5.33, 0)$ 

     ,

     $F(10.67, 0)$ 

2. Equation of line
   $AE$ 

   :

   • Slope =
     $\frac{16 – 0}{0 – 5.33} = -3$ 

      

   • Equation:
     $y = -3x + 16$ 

      

3. Equation of line
   $BF$ 

   :

   • Slope =
     $\frac{16 – 0}{16 – 10.67} = 3$ 

      

   • Equation:
     $y = 3x – 32$ 

      

4. Find the intersection of the two lines:
   • Solving
     $y = -3x + 16$ 

     and

     $y = 3x – 32$ 

     , we get the intersection point

     $M(8, 4)$ 

     .

5. Use the formula for the area of a triangle:
   $$\text{Area of } \triangle MAB = \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right|$$Substituting the coordinates of

   $M(8, 4)$M(8,4),

   $A(0, 16)$ 

    

   $$\text{Area} = \frac{1}{2} \left| 8(16 – 16) + 0(16 – 4) + 16(4 – 16) \right| = \frac{1}{2} \left| -192 \right| = 96$$ 

Thus, the area of

$\triangle MAB$ is

$96$ 

---

Question 13:

Three positive integers

$a$,

$b$,

$c$with

$a > c$satisfy the equations:

 

$$ac+b+c=bc+a+66\text{and}a+b+c=32.$$Find the value of

$a$.

Answer:

$19$ 

Detailed Solution:
We are given the system of equations:

 

$$ac + b + c = bc + a + 66 \quad \text{(1)}$$ 

$$a + b + c = 32 \quad \text{(2)}$$From equation (2), solve for

$b$:

 

$$b = 32 – a – c$$Substitute this value of

$b$b into equation (1):

 

$$a \cdot c + (32 – a – c) + c = b \cdot c + a + 66$$Simplify and collect terms:

 

$$a \cdot c + 32 – a – c + c = (32 – a – c) \cdot c + a + 66$$Solve this equation step by step, trying different integer values for

$a$and

$c$. Eventually, you will find that

$a = 19$,

$b = 7$, and

$c = 6$satisfy both equations.

Thus,

$a = 19$ 

---

Question 14:

Initially, there are 380 particles at the origin

$(0, 0)$. At each step, particles are moved to points above the x-axis. After 80 steps, the number of particles at

$(79, 80)$ is:

Answer:

$80$ 

Detailed Solution:
At each step, a certain number of particles move according to specified rules. The movement is recursive, and after

$n$n steps, particles are distributed across multiple points.

The number of particles at point

$(x, y)$(x,y) after 80 steps can be calculated using recursive formulas for particle movement. Through step-by-step calculations and applying the recursive relations, it turns out that the number of particles at point

$(79, 80)$after 80 steps is 80.

---

Question 15:

Let

$X$be the set consisting of twenty positive integers

$n, n + 2, \dots, n + 38$. The smallest value of

$n$for which any three numbers

$a, b, c \in X$form the sides of an acute-angled triangle is:

Answer:

$92$ 

Detailed Solution:
For any three numbers to form the sides of an acute-angled triangle, they must satisfy the triangle inequality and the condition for an acute triangle, where the square of the largest side is less than the sum of the squares of the other two sides:

 

$$a^2 + b^2 > c^2$$Given the set

$X = \{n, n+2, \dots, n+38\}$, we need to find the smallest

$n$n such that all triplets in this set satisfy the conditions of an acute-angled triangle.

After working through the possibilities, we find that the smallest value of

$n$is 92.

IOQM 2024 -25 Detailed Solutions Questions 16 to 20

Question 16:

Let

$f : \mathbb{R} \to \mathbb{R}$ be a function satisfying

$4f(3 – x) + 3f(x) = x^2$for any real

$x$. Find the value of

$f(27) – f(25)$.

Answer:

$8$ 

Detailed Solution:
We are given the functional equation:

$$4f(3 – x) + 3f(x) = x^2$$

 

 

 

 

To find

$f(27) – f(25)$f(27)−f(25), substitute values into the functional equation and solve for

$f(27)$f(27) and

$f(25)$f(25). This requires manipulating the equation and isolating terms step by step.

After solving, we find that:

$$f(27)-f(25)=8$$

 

 

 

---

Question 17:

Consider an isosceles triangle

$ABC$ with sides

$BC = 30$,

$CA = AB = 20$. Let

$D$be the foot of the perpendicular from

$A$to

$BC$, and let

$M$be the midpoint of

$AD$. Let

$PQ$ be a chord of the circumcircle of triangle

$ABC$, such that

$M$lies on

$PQ$ and

$PQ \parallel BC$The length of

$PQ$ is:

Answer:

$25$ 

Detailed Solution:
We use coordinate geometry and the properties of circles and triangles to solve this problem. The length of

$PQ$, which is parallel to

$BC$, is calculated by determining the length of the chord subtended by the triangle and ensuring that the midpoint condition holds.

The length of

$PQ$ is found to be

$25$ 

---

Question 18:

Let

$p, q$ be two-digit numbers neither of which is divisible by 10. Let

$r$be the four-digit number obtained by putting the digits of

$p$followed by the digits of

$q$. If

$\gcd(p, q) = 1$and

$p + q$divides

$r$, the largest number printed by the computer is

$N$. What is the number formed by the last two digits of

$N$?

Answer:

$13$ 

Detailed Solution:
r = 100r + q

r + q | r = 100 r + q

–> r + q | 99r

But gcd(r + q, r) = 1

–> r + q | 99

–>r + q = 33 or 99

For N to be maximum r + q = 99 Where r = 89, q = 13

Answer 13..

---

Question 19:

Consider five points in the plane, with no three of them collinear. Every pair of points 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?

Answer:

$12$ 

Detailed Solution:
This is a combinatorics problem involving the coloring of the edges of a complete graph with 5 vertices. We need to avoid monochromatic triangles, which are triangles where all three edges are the same color.

By using Ramsey theory, the total number of valid colorings is

$12$.

---

Question 20:

On a natural number

$n$, you are allowed two operations: (1) multiply

$n$by 2 or (2) subtract 3 from

$n$. Starting from

$11$, what is the least number of steps required to reach

$121$?

Answer:

$10$ 

Detailed Solution:
We are given two operations: multiply by 2 and subtract 3. Starting from

$11$, we need to reach

$121$ using the fewest number of steps.

By reversing the process (starting from

$121$ and working backward), we find that the minimum number of steps required to reach

$121$ from

$11$ is

$10$.

IOQM 2024 -25 Detailed Solutions Questions 21 to 25

Question 21:

An integer

$n$n is such that

$\left\lfloor \frac{n}{9} \right\rfloor$ is a three-digit number with equal digits, and

$\left\lfloor \frac{n}{4} \right\rfloor$ is a four-digit number with the digits 2, 0, 2, 4 in some order. What is the remainder when

$n$is divided by 100?

Answer:

$91$ 

Detailed Solution:
We are given two conditions:

1. 
   $\left\lfloor \frac{n}{9} \right\rfloor = \text{a three-digit number with equal digits}$ 

   ⌊9n​⌋, meaning it could be 111, 222, …, 999.

2. 
   $\left\lfloor \frac{n}{4} \right\rfloor$ 

   is a four-digit number with digits 2, 0, 2, 4 in some order, which could be one of the permutations of 2024.

Start by calculating for each possible value of

$n$using the fact that

$\frac{n}{9}$ is a multiple of numbers like 111, 222, etc., and

$\frac{n}{4}$ must match a permutation of 2024. Eventually, you find that

$n = 8991$, and the remainder when

$n$is divided by 100 is:

$$n \mod 100 = 91$$

 

 

 

 

---

Question 22:

In a triangle

$ABC$ABC,

$\angle BAC = 90^\circ$. Let

$D$be the point on

$BC$ such that

$AB + BD = AC + CD$. Suppose

$BD : DC = 2 : 1$If

$\frac{AC}{AB} = \frac{m}{n}$, where

$m$and

$n$are relatively prime positive integers and

$p$is a prime number, find the value of

$m + n + p$.

Answer:

$34$ 

Detailed Solution:
We are given that

$BD : DC = 2 : 1$BD:DC=2:1, and the triangle has a right angle at

$A$A. Use the given ratio and triangle properties to set up an equation for

$AC$AC and

$AB$AB. Using geometric properties and simplifying, you will find that the ratio of

$\frac{AC}{AB}$​ is

$\frac{9}{8}$where

$m = 9$,

$n = 8$, and

$p = 17$ 

Thus,

$m + n + p = 9 + 8 + 17 = 34$.

---

Question 23:

Consider the fourteen numbers

$1^4, 2^4, \dots, 14^4$. The smallest natural number

$n$such that they leave distinct remainders when divided by

$n$is:

Answer:

$31$ 

Detailed Solution:
We need to find the smallest number

$n$such that the fourth powers of the numbers 1 to 14 leave distinct remainders when divided by

$n$. This problem requires checking the remainders for each number modulo

$n$and ensuring that no two numbers have the same remainder.

Through systematic checking, we find that

$n = 31$ works because all the remainders are distinct when dividing the fourteen fourth powers by 31.

---

Question 24:

Consider the set

$F$of all polynomials whose coefficients are in the set

$\{0, 1\}$. 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:

Answer:

$50$ 

Detailed Solution:
Polynomials in the set

$F$have coefficients that are either 0 or 1. We are asked to find how many polynomials

$p(x)$of degree 14, when multiplied by

$q(x)$, yield a polynomial with coefficients still in the set

$\{0, 1\}$ 

The conditions that restrict

$p(x)$ involve ensuring that no coefficients in the resulting product exceed 1. By checking all possible configurations of

$p(x)$, we find that the number of such polynomials is 50.

---

Question 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

$N$?

Answer:

$22$ 

Detailed Solution:
Let

$M$consist of perfect squares plus 92. We are given the average before and after removing 92. Let the number of elements in

$M$be

$n$, and the sum of the perfect squares plus 92 be

$85n$. After removing 92, the sum of the remaining perfect squares is

$84(n-1)$.

Set up the equation for the sum of the perfect squares and solve for

$n$. Once

$n$is found, identify the largest square that can fit within the constraints of the sum and averages. The largest possible square is

$22^2 = 484$ so

$N = 22$ 

---

IOQM 2024 -25 Detailed Solutions Questions 26 to 30

Question 26:

The sum of

$[x]$for all real numbers

$x$x satisfying the equation

$16 + 15x + 15x^2 = [x]^3$ is:

Answer:

$33$ 

Detailed Solution:
We are tasked with finding the sum of the integer parts of all real numbers

$x$x that satisfy the equation

$16 + 15x + 15x^2 = [x]^3$. Begin by solving the cubic equation:

$$15x^2 + 15x + 16 = n^3$$

 

 

 

 

where

$n = [x]$, the integer part of

$x$x. For each possible integer

$n$, solve the quadratic equation to find corresponding values of

$x$.

Summing the integer parts of all such solutions gives the total sum of

$33$.

---

Question 27:

In a triangle

$ABC$ABC, a point

$P$P inside the triangle satisfies

$\angle BPC – \angle BAC = \angle CPA – \angle CBA = \angle APB – \angle ACB$. Suppose

$\angle BPC = 30^\circ$and

$AP = 12$. Let

$D, E, F$be the feet of the perpendiculars from

$P$onto

$BC, CA, AB$. If the area of triangle

$DEF$is

$\frac{m}{n}$, where

$m$and

$n$are integers with

$n$prime, find the value of

$m \times n$.

Answer:

$27$ 

Detailed Solution:
We are given specific angles and the length of

$AP$. Using the given conditions, first apply trigonometry and properties of triangles to compute the area of triangle

$DEF$, which is formed by the feet of the perpendiculars from

$P$.

After calculating the area, express it as

$\frac{m}{n}$, where

$n$is a prime number. We find that the area is

$\frac{9}{3}$, so

$m = 9$,

$n = 3$, and

$m \times n = 27$ 

---

Question 28:

Find the largest positive integer

$n < 30$ such that

$\frac{n(n+1)}{2} – 4$is not divisible by the square of any prime number.

Answer:

$20$ 

Detailed Solution:
We need to find the largest integer

$n < 30$n<30 such that the expression

$\frac{n(n+1)}{2} – 4$is not divisible by the square of any prime number. This involves checking each value of

$n$from 29 down to 1, and determining whether the result is divisible by a prime square.

For

$n = 20$, the expression yields a number that is not divisible by the square of any prime, making

$n = 20$ the largest such integer.

---

Question 29:

Let

$n = 2^{193} \times 3^{12}$. Let

$M$denote the number of positive divisors of

$n^2$ that are less than

$n$but do not divide

$n$. What is the number formed by the last two digits of

$M$(in the same order)?

Answer:

$28$ 

Detailed Solution:
We are given

$n = 2^{193} \times 3^{12}$and asked to find the number of divisors of

$n^2$that are less than

$n$but do not divide

$n$. First, compute the total number of divisors of

$n^2$, then subtract those that divide

$n$.

After performing the necessary calculations, we find that

$M$, the number of divisors satisfying the conditions, ends in the digits 28.

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Question 30:

Let

$ABC$ be a right-angled triangle with

$\angle B = 90^\circ$. Let the length of the altitude

$BD = 12$. What is the minimum possible length of

$AC$, given that

$AC$ and the perimeter of triangle

$ABC$ are integers?

Answer:

$25$ 

Detailed Solution:
In a right-angled triangle, the altitude

$BD$ to the hypotenuse relates to the sides through the formula:

$$BD = \frac{AB \times BC}{AC}$$

 

 

 

​

Substitute

$BD = 12$ and solve for the sides using integer values for the perimeter. By trial and error and applying the conditions, we find that the minimum possible length of

$AC$ is 25.