Math Challenge: The Odds and the Evens

Bon Crowder has recently been exploring the similarities between the addition of even and odd addends and the multiplication of positive and negative factors. Read her post on this topic to find what she has discovered.



I was taught to think of a negative number as a positive number times negative one; so, in this way the only real negative number is -1 and it is a direction vector. Similarly, i is the only real imaginary number, since all other imaginary numbers (not complex ones) are real numbers times i, the “imaginary vector”.

This is not a math challenge in the sense I have normally been applying the term. There are no computations, but there is a puzzle, one which I have yet to explore.

In Bon’s post, she connects the addition of odds and evens to the multiplication of negatives and positives. I wonder then if there is a way to express an odd number in terms of an even value times an “odd vector”? What would that odd vector be?


4 thoughts on “Math Challenge: The Odds and the Evens

  1. Note: I realize an easy “vector” is just to add one (+1) to the even number (2k) preceding the the odd number in question. And perhaps this is the only way to express an odd number (2k+1), but +1 does not have the feel of a vector. Is there a vector we can multiply to an even number to get the corresponding odd number or are we stuck with +1?

    One difference of -1 and i as vectors from any possible oddness vector is that the real (absolute) value remains the same when they are applied. This is not true for an even to odd conversion. I guess that means the oddness vector must be an addition of one.

    So what do we do when our students run down similar blind alleys? Do we learn anything from this exercise at all?


  2. On August 18, James Tanton tweeted this problem:

    Let S be a set of positive integers such that every number, except 1, has an even number of factors appearing in S. What is S?

    Bon Crowder responded with this video in which she proposed that S was the set of all numbers that did not have common prime factors, such as two 4s, in them.

    However, it occurs to me that if a number has two unique common prime factors, such as two 4s and two 5s, then it belongs in S, since the number once again has an even number of unique prime factors.

    Taking this further, I propose that if a number has an odd number of common prime factors, it must not belong in S; all other numbers do.

    So, why am I bringing this up here?

    Have you read Bon’s post and mine about odd negatives and odd odds?

    I love discovering connections and patterns in math.

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