# Properties of even and odd numbers

## Problem

There's a running total that starts at 0. Two players alternate taking turns. Each turn consists of adding 1, 3, 5, or 7 to the running total. The goal is to get the running total to 37. Anything more and it's a tie. What's the maximum number of moves possible? the minimum? During your trip to 37, can all of the numbers, 1, 3, 5, and 7, be used? Does it matter who goes first, and if so, why?

## Solution

The maximum number of moves is obviously 37. The minimum number of moves is 7, and can be achieved in 2 ways

\begin{align} & 7 + 7 + 7 + 7 + 7 + 1 + 1 \\ & 7 + 7 + 7 + 7 + 5 + 3 + 1 \end{align}

The first player always wins, because the running total alternates between even and odd. The running total starts at 0, which is even. Then the first player adds an odd number to it. even + odd = odd, so the running total is now odd. Then the second player adds an odd number. odd + odd = even, so the running total is now even. Continuing this way, the first player can only reach odd numbers, and the second can only reach even numbers. Because 37 is odd, the first player will always win.

Source

The following puzzle can be given immediately after students have learned that odd + even = odd.

Draw four circles, such that each circle contains an odd number of roses, each circle contains a different number of roses, and no two circles touch or intersect:

Here's the solution:

Source: Test Your Math IQ by Steve Ryan

This puzzle could be modified in various ways. For example, we could make a puzzle where not all circles have the same center. We could make a puzzle where not all circles are concentric. We could lift the restriction that circles may not intersect. We could use squares, or some other shape, instead of circles. If I were to make such a puzzle I would use dots instead of roses, as then it's easier to see which dots would be bound by the solution being imagined. This puzzle could also be adapted to work with other number types. For example, we could ask for each circle to contain a different triangular number, or prime number. Given the coordinates, this problem could be an interesting exercise for programmers. See the smallest-circle problem if this interests you.

Choose four distinct digits to put in the squares so that the difference between adjacent squares is odd. What can be said about the sum of adjacent squares? Next, choose four distinct digits so that the difference between adjacent squares is even. Now what can be said about the sum of adjacent squares? Solution. What if there were 3 squares? 5? n? Solution.

Problem:

There are 17 people from Neptune, each of which, have either 5 or 3 hands. To clarify, it's possible that some have 5 hands, while others have 3 hands. There are 13 people from Saturn, each of which have 2 hands. If all of these people were to meet, could they all hold hands, such that no hand is empty?

Solution

No. If we have odd number of addends, and each addend is odd, then the sum will be odd. Thus, the 17 people from Neptune, each of which, have either 5 or 3 hands, gives us an odd number of hands. If we have an odd number of addends, and each adden is even, then the sum will be even. Thus, the 13 people from Saturn, each of which have 2 hands, gives us an even number of hands. But odd + even = odd, and thus, there are an odd number of hands in total. Hence, one person's hand must be empty.

Source: Adapted heavily from "The Inquisitive Problem Solver, Problem 18."