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...

For how many natural numbers n between 1 and 2014 (both inclusive) is \(\frac{8n}{9999-n}\) an integer? Correct Answer : 1. Solution: We are given that \(\frac{8n}{9999-n}\) is an integer. We can rewrite the expression as follows: $$ \frac{8n}{9999-n} = \frac{8}{(9999/n - 1)} $$ Solution: \(\frac{8}{9999-n} = \frac{8}{\frac{9999}{n}-1} = integer\) For integer value \((\frac{9999}{n}-1)\) is a factor of 8. So possible values of \((\frac{9999}{n}-1) = 1, 2, 4, 8\) \(\frac{9999}{n} = 2, 3, 5, 9\) \(n = \frac{9999}{2}, \frac{9999}{3}, \frac{9999}{5}, \frac{9999}{9}\) Integer \(n = 3333, 1111\) But n So only possible value of n = 1111. So 1 is correct answer.