Difference between revisions of "1998 USAMO Problems/Problem 1"
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== Solution == | == Solution == | ||
− | + | Notice that <math>|a_i - b_i| \equiv 1 \pmod 5</math>, so <math>S=|a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}| \equiv 1+1+\cdots + 1 \equiv 999 \equiv 4 \bmod{5}</math>. | |
− | + | Also, for integers M, N we have <math>|M-N| \equiv M-N \equiv M+N \bmod{2}</math>. | |
− | + | Thus, we also have <math>S \equiv a_1+b_1+a_2+b_2+\cdots +a_{999}+b_{999} \equiv 1+2+ \cdots +1998 \equiv 999*1999 \equiv 1 \bmod{2}</math> also, so by the Chinese Remainder Theorem <math>S \equiv 9\bmod{10}</math>. Thus, <math>S</math> ends in the digit 9, as desired. | |
==See Also== | ==See Also== |
Revision as of 23:08, 12 February 2015
Problem
Suppose that the set has been partitioned into disjoint pairs () so that for all , equals or . Prove that the sum ends in the digit .
Solution
Notice that , so .
Also, for integers M, N we have .
Thus, we also have also, so by the Chinese Remainder Theorem . Thus, ends in the digit 9, as desired.
See Also
1998 USAMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.