There are forty weights: $1, 2, \cdots , 40$ grams. Ten weights with even masses were put on the left pan of a balance. Ten weights with odd masses were put on the right pan of the balance. The left and the right pans are balanced. Prove that one pan contains two weights whose masses differ by exactly $20$ grams. (4 points)
Problem
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Tags: modular arithmetic
andersonw
03.09.2010 23:35
Assume for contradiction that neither pan contains two weights whose masses differ by exactly 20 grams.
Split up the even weights into the sets $\{2, 22\}, \{4, 24\}, \cdots, \{20, 40\}$. There are ten sets, and at most one weight from each set may be picked, so we must pick exactly one weight from each set.
Similarly, we must also pick exactly one weight from each of $\{1, 21\}, \{3, 23\}, \cdots, \{19, 39\}$.
Now, consider the sum of each side mod 20. The left side has sum $2(2+4+6+8+10)\equiv 0 \pmod{20}$.
The right side has sum $2(1+3+5+7+9)=10\pmod{20}$. As 0 and 10 are not equal, we have reached a contradiction and we are done.
Takeya.O
05.07.2016 17:06
Hello, andersonw. I think that left side sum is 2＋...＋20≡10 mod 20, and right side sum is 1＋...＋19≡0 mod 20.