Answer:

The answer is no. Suppose that the first gear rotates clockwise. Then the second gear MUST rotate counter-clockwise, the third clockwise again, the FOURTH counterclockwise, and so on. It is clear that the ”odd” gears must rotate clockwise, while the ”even” gears must rotate counter-clockwise. But then the first and eleventh gears must rotate in the same direction. This is a contradiction. The main idea in the solution to this problem is that the gears rotating clockwise and counter-clockwise alternate. Similarly, for 572 gears, the rotation is possible.

2. (6 points) Prove that if a number has an odd number of divisors, then it is a perfect square.

Solution: If a number n is not a perfect square, its divisors can be paired up (all such pairs will have a number > 

√ n and a number <  n), thus making the number of √

divisors even. However, in CASE of odd number of divisors, we have an extra DIVISOR, which is possible only if we have a square number.

3. (8 points) If p, 4p2 + 1, and 6p2 + 1 are prime numbers, find p. Solution: The answer is p = 5. Analyze the remainders upon division by 5.

4. (10 points) How many ways are there to place: a) two bishops, b) two knights, c) two queens on a chessboard so that they do not attack each other? Solution:

Stamatics, IIT Kanpur Page 1 of 5Mathemania-1 Solutions

5. (12 points) Prove that all the numbers in the series 10001, 100010001, 1000100010001, ...

are composite.