1180.61 – Mathies and Websters


A cool group is reviving an old library of high school math problems and building a new website for them. The group consists of Mathies and Websters. One Mathie says, "There are as many Mathies besides me in this very cool group as there are Websters." A Webster says, "There are twice as many Mathies as other Websters." How many Mathies and Websters are there?

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Solution

Trial and error with small numbers will show fairly quickly that there are 4 Mathies and 3 Websters. Algebra-minded students may prefer a more formal solution:

Suppose there are mm Mathies and ww Websters (total). A Mathie has m 1m - 1 fellow Mathies. The Mathie who spoke claimed, m 1= w or m = w +1.m - 1 = w \text{ or } m = w + 1. A Webster has w 1w - 1 fellow Websters. The Webster who spoke said, m =2(w1),m = 2(w - 1), or m=2w 2.m = 2w - 2. So 2w2= w +1,2w - 2 = w + 1, and thus w =3w = 3 and m =4m = 4.

(Historical note. This happens to be almost exactly the number of folks who did the initial work on this website.)