Monthly Math Contest

Here is your chance to have a little fun and get a little recognition! Each month, the American Society of Mathematics sends a new problem to Dr. Koo. The first student to submit a correct answer will be the winner for the month, and have their name posted on the Department of Mathematical Sciences web page. The top problem solver for the year will be invited to attend the US National Collegiate Mathematics Championship in Madison, Wisconsin.

Rules:

Write your solution clearly and show all work.

Take your solution to the Secretary, Math Office Room 227,with name, date and time.

Include your name, contact information, date and time of submission.

The first correct solution will be the winner for the month.

All correct solutions will be recognized. Names posted on Math- Sciences Webpage.

A new problem will be posted the first week of each month during the spring and fall semesters.

The top problem solver for the year will be invited to attend the US National Collegiate Mathematics Championship, to be held in Madison, Wisconsin 2008.

Questions relating to this competition can be directed to Reginald Koo, at regk@usca.edu.

Problem October 2009:

Determine all positive integers m and n such that (m + n)² = (m + 1)² + (n + 1)².

Deadline is Friday October 30, at 4.00 pm.

Problem September 2009:

For f(x) = x³+6x²-15x+k, the absolute maximum and absolute minimum values on the interval [-10, 2] have the same absolute value. Find the value of k.

Deadline is September 30, at 4.00 pm.

Problem March 2009:

(i) Which of the five numbers 2007, 2008, 2009, 2010 and 2011 has the largest number of (positive) factors, and which one has the fewest number of factors?
[For example, 20 has six factors, namely, 1, 2, 4, 5, 10, 20.]

(ii) Determine the number of factors of the number 2007 . 2008 . 2009 . 2010 . 2011: Submit solutions to Math Dept Administrative Assistant; include date and time submitted.

Deadline is Friday March 27, at 4.00 pm.

Problem Feb 2009:

Submit solutions to Math Dept secretary, include date and time submitted. Deadline is Friday February 27, at 4.00 pm.

Winners: First Place: Chris Edgar

Problem Nov 2008:

Winners: First Place: Lee Chi Heng Others: Mark Woodhams

Problem October 2008:

Problem. How many of the positive factors of the number 36,000,000 are not perfect squares?

Problem September 2008:

Evaluate

Submit solutions (show all work) to Math Dept secretary. Include date and time submitted. Deadline is Monday September 22, 2008 at 4.00 pm.

Answer:

Winners: First Place: Chris Edger

Problem April 2008

Find the value of the sum
Submit solutions to Math Dept secretary, include date and time submitted. Deadline is Wednesday April 30, 2008 at 4.00 pm.

Winners: First Place: Matthew Coffin

Problem March 2008

Derivative Calculate the derivative with respect to x of the function y = x^{x ^x}.
Deadline for solutions is by March 31, 2008 at 4.00 p.m.

Winners: First Place: Behzad Torkian

Problem February 2008

A logarithmic equation. Solve the equation (ln x)² − 2.5(ln x)(ln(4x − 5)) + (ln(4x − 5))² = 0, where x and all expressions in the equation are real.
Deadline for solutions is by Feb 28, 2008 at 4.00 p.m.

Problem November 2007

Consider the number 12355699. If we write each of the digits in this number on separate slips of paper, put them in a bowl, and draw three of the numbers at random, without replacement, what is the probability that the sum of the numbers drawn will be even?
Deadline for submission of solution: 30 Nov at 4.00 p.m.

October Problem, Answer and Winner

October 2007 PROBLEM : 6 DVDS Julio has 6 favorite DVDs. If each DVD is 20 minutes long and Julio wants to watch all 6 DVDs over and over in every conceivable order, how long (in hours) will it take him to accomplish this? He will, for example, watch all 6 in a row in one order and then watch the same 6 in a different order, repeating this until all possible orders are achieved.

Solution to October 2007 Problem. There are 6! = 720 different orders of the DVDs. Each order requires 6·20 = 120 minutes or 2 hours to watch. Watching all possible orders then requires 720 · 2 = 1440 hours.

Winners: First Place: Brian Staples Other correct solutions: Matthew Coffin, Mark Woodhams.