# A cool coincidence with the 2014 Putnam and an old blog post

A few weeks ago I wrote up a fun exercise that my younger son and I had worked through:

That exercise explored a fairly straightforward situation: what two digit numbers are equal to 4 times the sum of their digits? There are a few: 12, 24, 36, and 48. My son saw the pattern that the first digit had to be twice the second digit, so I asked him: what about the number Fifty Ten?

My question is sort of silly, I know, but I wanted to get him thinking about place value. Fifty Ten would be the same as 60, but the sum of the digits of these two “numbers” is different.

This morning I got quite a surprise reading problem B1 from the 2014 Putnam Exam:

The 2014 Putnam exam at the Art of Problem Solving site

Essentially the question asks about this situation: if you allow 10 to be a digit in base 10, some positive integers will have a unique representation in the new number system and some won’t. For example, the number ten could be written in the usual way as 10, or in a new way as the single “digit” (10). The number 19, however, can only be written one way in this new number system. What are the positive integers will have a unique representation in this new number system?

What a cool coincidence – a question where fifty ten is actually a number!

How fun to have an old project with my younger son sort of overlap with a Putnam question 🙂

Updated to include talking through the problem with my son: