# Difference between revisions of "1998 USAMO Problems/Problem 1"

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+ | 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 99 \equiv 4 (mod 5)</math>. | ||

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+ | For integers M, N we have <math>|M-N| \equiv M-N \equiv M+N (mod 2)</math>. | ||

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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>. |

## Revision as of 10:40, 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

*This problem needs a solution. If you have a solution for it, please help us out by adding it.*

If , then .

For integers M, N we have .

So we also have also, so .