Annotated Solution to the Jotto Puzzle of the Day

May 7, 2003

farer - 2
solar - 2
means - 2
muley - 1
sheth - 1
flews - 2
renew - 1
Hmm, not an obvious place to start. renew has a repeated letter and a score of 1, so that breaks down into only 4 possibilities. Let's start there. e is the most common letter, and it's also shared by a lot of the other words, so we'll start by assuming there's an e. This eliminates r, n, and w, so let's reduce from there:


Reducing with the new zeros gives us:


af gives us a contradiction (fa=1), so we know there is no e

. Let's try to r next. Reducing gives us:


We know there can't be both an a and an s (sola=1), so there must be an m (mas=2). Reducing gives us:


Since (as=1), (sola=1) tells us that there can't be an o or an l (since the 1 in (sola=1) has to come from either the a or the s). With this information and the new zero, we reduce again to get:


Let's see what we get when we assume an a:


That's 6 letters, so a is out. That means it has to be the s (as=1). Let's reduce:


We now have two choices. We can exclude rmsrl because it has no vowels. That leaves us with rmsf. A quick look at our original words tells us that we've seen every vowel except i. Adding that in gives us rmsfi. A little anagramming gives us the solution: firms.

Jotto Puzzle of the Day
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Last modified: 08 May 2003