## Five through Fifteen

*April 4, 2019 by Chris Johnson. Filed under math, public.*

On our walks home from school, my 8-year-old son has been telling me of a math problem that he has not been able to solve. Yesterday, after a week of not finding a solution, he wrote down the problem on paper for us to investigate together. Here is the task in my own words:

Distribute all of the numbers in [5, 15] into boxes A and B. The numbers in box A must sum up to an even number, while the numbers in box B must sum up to an odd number.

My son had been trying to solve this by generating random distributions, but he had not been able to make it work. I suggested that we step back a second to see what must be true of each box. Here’s what we reasoned:

- The even numbers have absolutely no impact on the boxes. If a box’s sum is even, adding an even number to it will keep it even. If a box’s sum is odd, adding an even number will keep it odd. Because the even numbers have no impact, we can distribute them how we please. Therefore we will concern ourselves only with the distribution of the odd numbers.
- Two odds added together form an even number. Three odds added together form an odd number. Four odds added together form an even number. And so on. In general, an odd number of odds sums to an odd number, while an even number of odds sums to an even.
- Based on statement 2, for box A to contain an even sum, it must contain an even number of odd numbers.
- Based on statement 2, for box B to contain an odd sum, it must contain an odd number of odd numbers.
- Based on statements 3 and 4, there is an even number of odds in box A and an odd number of odds in box B, and since an even plus an odd is odd, there must be in total an odd number of odd numbers.
- But there are an even six odd numbers in our range: 5, 7, 9, 11, 13, and 15. There’s no way we can achieve statements 3 and 4.

After school today we stopped in and asked my son’s teacher if we were misinterpreting the problem statement somewhere. I tried to walk through the proof that there was no distribution, but I failed to convince her. Probably because I didn’t want to be one of *those* parents. Or probably because phrases like “an even number of odds” get really disorienting.

The teacher was kind enough to check the solution manual for us. But she found that the provided answer incorrectly had even sums in both boxes. My son and I felt vindicated. She apologized to my son for all the time he spent. We did not accept her apology as the unsolvable problem was more fun than a solvable one.

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