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In Search of the Magic Pair

Zach has chosen five numbers from the set {1, 2, 3, 4, 5, 6, 7}. If he told Claudia what the product of the chosen numbers was, that would not be enough information for Claudia to figure out whether the sum of the chosen numbers was even or odd. What is the product of the chosen numbers?

Essentially, we want to find distinct classes of 5 numbers from \{1,2,3,4,5,6,7\} whose products are the same but the parity of whose sum are not. One way of doing this is to look at the 7 \choose 5 ways of multiplying 5 numbers from the set and record the equivalence classes that arise. However, by giving the product of 5 numbers, an equivalent problem is to find which two numbers were not chosen. Hence, a more manageable technique is to just look at the {7 \choose 2}={7\choose 5} = 21 ways of choosing the two leftover numbers.

Let’s find all the ways of choosing two numbers from the set such that their products are the same. A quick glance gives us the classes \{(2,3),(1,6)\} and \{(3,4),(2,6)\} which can be shown via exhaustive enumeration to be the only satisfying pairs.

Now that we only have two cases left. Our answer is either \{(1,2,5,6,7),(1,3,4,5,7)\}_{420} or \{(1,4,5,6,7),(2,3,4,5,7)\}_{840}.

Because both 1+4+5+6+7 and 2,3,4,5,7 are odd, we can already eliminate 840 from the possible answers. This just leaves us with 420. Let’s make sure that 420 is the answer:

    \begin{align*} 1+2+5+6+7 &= 21 \\ 1+3+4+5+7 &= 20 \end{align*}

Since one is odd and the other is even, 420 is our answer.

Posted By Lee