This problem is not original, but pushing the solution in this way is fun. Watch, or play along. See if you can figure out what a nice ending might be.

Find a number that is 1 more than its reciprocal

Make sure the audience knows what a reciprocal is:  It’s what you multiply a number by to get 1 as the product, or colloquially, the “flip” of a number. For example the reciprocal of 9 is $\frac{1}{9}$. The reciprocal of $\frac{7}{8}$ is $\frac{8}{7}$.

I usually allow kids to explore in any direction, but for today’s purposes I don’t want that to happen. I’m going to control the investigation.

Get the kids to make some naïve guesses: 1 is too small since the reciprocal of 1 is 1. 2 is too big since the reciprocal of 2 is $\frac{1}{2}$ (and 2 – $\frac{1}{2}$ = $\frac{3}{2}$).

So let’s start more serious guessing. $\frac{3}{2}$$\frac{2}{3}$ = $\frac{5}{6}$, so that’s wrong. But it’s not such a terrible guess. $\frac{5}{6}$ is only a little less than one. We need a slightly bigger number. Add 1 to $\frac{2}{3}$. That will give us something a little bigger than $\frac{3}{2}$.

$\frac{2}{3}$ + 1 = $\frac{5}{3}$. Did that just give us the solution? No. $\frac{5}{3}$$\frac{3}{5}$ = $\frac{16}{15}$,  a little more than 1, but very little more than 1. But that helps us get the next guess. If we add 1 to $\frac{3}{5}$ we will get something very slightly less than $\frac{5}{3}$.

$\frac{3}{5}$ + 1 = can you imagine where I might be leading the students? What would you like the students to notice? What concepts would you like to share with them?

If you comment, mention the age of the children you imagine working with. I’ve been speaking with a few teachers, and we are trying versions of this with 10 year olds, 17 year olds, and everything in between….

I haven’t posted a mathematics puzzle or problem (for kids or adults) in quite some time. I hope there’s someone out there who still likes this.