S52 - In the contrary case all of the S k have a non-zero remainder when divided by 1998 By the pigeonhole principle(there are 1998 sums but only

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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.
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Unformatted text preview: In the contrary case, all of the S k have a non-zero 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....
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This note was uploaded on 03/01/2010 for the course MATH 301 taught by Professor Albertodelgado during the Spring '10 term at Bradley.

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S52 - In the contrary case all of the S k have a non-zero remainder when divided by 1998 By the pigeonhole principle(there are 1998 sums but only

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