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

Revision as of 08:02, 13 September 2012

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

If $S=|a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}|$ , then $S \equiv 1+1+\cdots + 1 \equiv 999 \equiv 4 \bmod{5}$ .

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

So 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 $S \equiv 9\bmod{10}$ .

@@ Line 4: / Line 4: @@
 == 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 999 \equiv 4 (mod 5)</math>.
+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 \bmod{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 \bmod{2}</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>.
+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 \bmod{2}</math> also, so <math>S \equiv 9\bmod{10}</math>.
 ==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

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

Revision as of 08:02, 13 September 2012

Problem

Solution

See Also