Math IOQM

Q14(combinatorics) . 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: Original problem has large numbers like 3^80. So let's see how to solve a smaller problem. Let there be 3^2 particles with the same rules. Let's figure out what happens after 2 steps. Initially 9 particles at (0,0) First step: (0,1): 3 particles (1,1): 3 particle...

Q1 (number-theory). The smallest positive integer that does not divide 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 is: Solution : This product has all the integers from 1 to 9. So smallest integer not dividing it would be the next prime after 9. Which is 11. Answer: 11 Q2 (combinatorics): The number of four-digit odd numbers having digits 1, 2, 3, 4, each occurring exactly once, is: Solution: Last digit has to be 1 or 3: 2C1 Choosing remaining 3 digits: 3C3 Arranging those 3 digits: 3! Answer: 2C1*3C3*3! = 12 Q3 (number-theory): The number obtained by taking the last two digits of \(5^{2024}\) in the same order is: Solution: To get last 2 digits from number, divide by 100 and take the remainder. So we have to compute mod 100. 25 mod 100 = 25 25.5 mod 100 = 125 mod 100 = 25 5^3.5 mod 100 = 25.5 mod 100 = 25 5^4.5 mod 100 = 25.5 mod 100 = 25 ..... So: 5^2024 mod 100 = 25. Q4 (geometry): Let ABCD be a quadrilateral with \(\angle ADC = 70^\circ\), \(\angle ACD = 70^\circ\), \(\angle ACB = 10^\circ\...