The Department of Mathematics invites all Catholic University students to compete, for the fun of it, in a mathematics contest. The contest consists of mathematical problems or puzzles which can be understood by anyone with the usual high school mathematics background. The most successful contestants will be invited to the Mathematics Department end-of-semester party to receive prizes. There will be prizes for the students who solve the most problems and for those who submit the most interesting or original solutions (even if for only one problem).

For Spring semester 2021, see the questions below, and the JPG image at the bottom of the page. Send your solutions by April 26, 2021 to Dr. Alexander Levin at the Mathematics Department, levin@cua.edu. They need not be typed but should be legible and should show or explain how you solved the puzzle.

**Problem 1**.With how many zeros the number 300! = 1 *· *2 *· *3 *· *4 *· · · · · *299 *· *300 ends? Justify your answer. (The product of all positive integers from 1 to a positive integer *n *is called *n *factorial and is denoted by *n*! .)

**Problem 2**. “This might interest you, Chris,” said Paul after a university faculty meeting. “My age and the ages of each of my three nephews are all prime numbers, and the sum of our ages is 50.” “Now,” said Chris, who knew Paul’s age, “I can tell you the ages of your three nephews.”

Even though you don’t know Paul’s age, can you tell the ages of his nephews? (Recall that a prime number is a positive integer *p > *1 that has exactly two positive integers divisors, 1 and *p*.)

**Problem 3**. Find the largest group of different positive integers less than 80 such that no combination of them added together totals 80.

**Problem 4**. In the multiplication problem

2 *· ab · cde *= *fghi, *

each of the ten digits is different. (Thus, one of the digits is 2 and *a, b, c, d, e, f, g, h, i *stand for distinct digits other than 2.) Find the values of *a, b, c, d, e, f, g, h, i*.

**Problem 5**. If a line from vertex *C *of a triangle *ABC *bisects the median from *A*, prove that it divides the side *AB *into ratio 1 : 2.

**Problem 6**. Find all real solutions of the equation

(12*x − *1)(6*x − *1)(4*x − *1)(3*x − *1) = 1*. *

(That is, find all real numbers that satisfy the equation.)