Difference between revisions of "1998 USAMO Problems/Problem 1"
(added box) |
(→Solution) |
||
Line 4: | Line 4: | ||
== Solution == | == Solution == | ||
− | If <math>S=|a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}|</math>, then <math>S \equiv 1+1+\cdots + 1 \equiv | + | If <math>S=|a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}|</math>, then <math>S \equiv 1+1+\cdots + 1 \equiv 999 \equiv 4 (mod 5)</math>. |
For integers M, N we have <math>|M-N| \equiv M-N \equiv M+N (mod 2)</math>. | For integers M, N we have <math>|M-N| \equiv M-N \equiv M+N (mod 2)</math>. |
Revision as of 14:34, 7 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 |