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Problem of the Month

Here is a problem for the end of summer. Deadline: Aug. 31, 2002

A person takes at least one aspirin a day for 30 days. 
Suppose he takes 45 aspirin altogether. Is it possible that 
in some sequence of consecutive days, he takes exactly 14 aspirin?

Problem-Solving Contest Rules and Instructions

  1. Contestants must be undergraduates at NIU.
  2. Solutions may be solved and submitted by groups.
  3. Write your solutions clearly and show all of your work.
  4. Include your name(s), address(es), and phone number(s).
Please submit your response by [campus] mail to
Math Club
Northern Illinois University
Department of Mathematical Sciences
C/O Dr Blecksmith
DeKalb, IL 60115

or email to richard@math.niu.edu

Be sure to include your name, address, and telephone number.

The names of those who submit correct solutions will be posted on the Math Club bulletin board outside of the Math Department office and on the web page of the NIU Math Club. Small prizes may be awarded for correct solutions!

Solution to last month's problem:

We were asked to show that if p > 5 is a prime number, then there exist positive integers a, b such that p-a2 is a proper divisor of p-b2.

(A proper divisor of an integer n is a divisor which is greater than 1 and less than n.)

Solution Interestingly, p doesn't have to be prime, exactly, but not any old integer will work. Here's one way to see what will work.

Write p as a2 + k with integers a and k, where k > 0 is chosen to be as small as possible. Then let b = |a - k|.

Then p - b2 = p - (a - k)2 = k + 2ak - k2, which is a multiple of k = p - a2. This shows that p-a2 is a divisor of p-b2; we need only check that it is a proper divisor. But the two can't be equal because b = |a - k| is clearly strictly smaller than a (Exercise: what did we assume that makes k nonzero? Also, why can't it be true that p-a2=1? Try replacing b with composite numbers, up to 100 say, and see what happens...)

Solutions to prior months' problems:

We were asked to show that for a, b, c > 0, we have

1/(a3+b3+abc) + 1/(b3+c3+abc) + 1/(c3+a3+abc) <= 1/abc

Well for positive numbers a and b, we know a+b is positive and thus (a-b)2(a+b) is non-negative, or, stated a little differently,

(1) (a - b)(a2 - b2) >= 0.
Expanding this out and rearranging terms, we deduce that a3 + b3 >= ab(a+b). It follows that : 1/(a3 + b3 + abc) <= 1/(ab(a+b) + abc) = c/(abc(a+b+c)) and we have similar inequalities for the other terms. Summing, we get the desired inequality.

Note that equality appears iff there are equalities in (1), that is, iff a = b = c.

Director of Undergraduate Studies: Dave Rusin
E-mail: ugradprog@math.niu.edu [HOME] Back to the NIU Math Department Web page
Last modified: 2001/12/23 (djrr)