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2000 SMT/Advanced Topics Problems/Problem 4

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Problem

Five positive integers from $1$ to $15$ are chosen without replacement. What is the probability that their sum is divisible by $3?$


SMT Solution

The possibilities for the numbers are: 

   all five are divisible by $3$
   three are divisible by $3,$ one is $\equiv 1 \pmod 3$ and one is $\equiv 2\pmod 3$
   two are divisible by $3,$ and the other three are either $\equiv 1\pmod 3$ or $\equiv 2\pmod 3$
   one is divisible by $3,$ two are $\equiv 1\pmod 3$ and two are $\equiv 2\pmod 3$
   four are $\equiv 1\pmod 3$ and one is $\equiv 2\pmod 3$
   four are $\equiv 2\pmod 3$ and one is $\equiv 1\pmod 3$

This gives us $1001$ possible combinations out of $\binom{15}{5}$ or $3003$. So, the probability is $\frac{1001}{3003}=\mathbf{1}{3}.$




Credit

Problem and solution were taken from https://sumo.stanford.edu/old/smt/2000/

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