Difference between revisions of "1998 USAMO Problems/Problem 1"
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So 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 (mod 2)</math> also, so <math>S \equiv 9 (mod 10)</math>. | So 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 (mod 2)</math> also, so <math>S \equiv 9 (mod 10)</math>. | ||
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+ | ==See Also== | ||
+ | {{USAMO newbox|year=1998|before=First Problem|num-a=2}} |
Revision as of 11:44, 6 June 2011
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
If , then .
For integers M, N we have .
So we also have also, so .
See Also
1998 USAMO (Problems • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |