1998 USAMO Problems/Problem 1

Problem

Suppose that the set $\{1,2,\cdots, 1998\}$ has been partitioned into disjoint pairs $\{a_i,b_i\}$ ( $1\leq i\leq 999$ ) so that for all $i$ , $|a_i-b_i|$ equals $1$ or $6$ . Prove that the sum $|a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}|$ ends in the digit $9$ .

Solution

Notice that $|a_i - b_i| \equiv 1 \pmod 5$ , so $S=|a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}| \equiv 1+1+\cdots + 1 \equiv 999 \equiv 4 \bmod{5}$ .

Also, for integers M, N we have $|M-N| \equiv M-N \equiv M+N \bmod{2}$ .

Thus, we also have $S \equiv a_1+b_1+a_2+b_2+\cdots +a_{999}+b_{999} \equiv 1+2+ \cdots +1998 \equiv 999*1999 \equiv 1 \bmod{2}$ also, so by the Chinese Remainder Theorem $S \equiv 9\bmod{10}$ . Thus, $S$ ends in the digit 9, as desired.

1998 USAMO (Problems • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6
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1998 USAMO Problems/Problem 1

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