This preview shows pages 1–2. Sign up to view the full content.
Solution to Problem 52
There were no solutions submitted from Bradley University students  it was the week of final exams,
what was I thinking! Correct solutions were received from Steve Young, Tim Kelley, CWLDOC,
Aaron Kahn, Terauchi Kimio.
Let
A
1
,...,
A
1998
be the numbers selected, and let
S
k
=
A
1
+ .
.. +
A
k
.
If any one of the
S
k
is divisible by 1998, we’re done.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
Unformatted text preview: In the contrary case, all of the S k have a nonzero remainder when divided by 1998. By the pigeonhole principle (there are 1998 sums but only 1997 possible remainder  1 to 1997) two of the sums must have the same remainder upon division by 1998. The difference of these two sums is divisible by 1998. this page....
View
Full
Document
This note was uploaded on 03/01/2010 for the course MATH 301 taught by Professor Albertodelgado during the Spring '10 term at Bradley.
 Spring '10
 AlbertoDelgado
 Combinatorics

Click to edit the document details