Instead of a weekend challenge question last week, I posted a question writing contest that I thought would be a big hit, but which has thus far generated much enthusiastic sound and fury, and only one actual question. Ah well. I maintain my position that if you are able to synthesize good SAT questions, you’ll be in very good shape to do well on test day. But if you don’t want to try it, I won’t shed any additional* tears.

Anyway, a return to form: This weekend challenge is designed to frustrate, and then eventually satisfy. The prize for the first correct response will be Beta Access to my Math Guide.

A precocious toddler is making a pattern with his alphabet blocks, each of which displays a letter (A-Z) in one of 5 colors: red, blue, green, orange, yellow. He has a LOT of blocks to work with, so he sets about putting them in alphabetical order, and also color order. The pattern goes as follows: Red A, Blue B, Green C, Orange D, Yellow E, Red F, etc. When he gets to the end of the alphabet, he starts at the beginning again. Assuming he doesn’t screw up the pattern, the

n^{th}block will be his first Green N. What isn?

Put your answers in the comments, and I’ll post the solution Monday. Good luck!

UPDATE: Nice work, Robin. I hope you enjoy the Beta!

Solution below the cut.

This is basically a 2-for-the-price-of-one pattern question. I’m going to explain it the snappiest way I know how, but you should also see Robin’s winning comments for a tidy little solution if you don’t like mine.

1) Since there are 5 colors (and we use a base 10 counting system) it’s going to be easy to track the green blocks. The first green block will be the 3rd block, and then the 8th block, and then the 13th, and then the 18th, etc. Basically, if a term has a units digit of 3 or 8, it’s a green block.

2) So now all you need to do is figure out when is the first time the letter N will fall on one of those terms. N is the 14th letter, so no go there. There are 26 letters, so the next time N comes up will be the 14+26 = 40th term. Then the 66th, 92nd, and ** 118th**. 118 ends in 8,

*SO THE 118th BLOCK IS GREEN AS HELL!*Yessssss.

* I’ve been sobbing all week. You monsters.

## Comments (8)

n = 118 …

n = 118 because:

118 divided by 26 has a remainder of 14. Letter “n” is the 14th letter of the alphabet. When 118 is divided by 5, the remainder is 3, and green is the third color in the sequence.

This is zengc. This time I have signed in using my Google account to post my answer since anon. posts don’t count (I looked at your earlier challenges).

n = 118.

Enjoy the Beta!

No holds barred right? Well, here’s my method 🙂 …

http://min.us/lhLAJJzSz0G32

Not that this is by any means the most efficient way, but could you use modular arithmetic to solve this problem?

I should think so. I’m only going on what I just learned about modular arithmetic on wikipedia, but I’d have to imagine this would be a fairly elementary application of it for someone who’s studied it. Care to try?

The way I tried to solve it in modular arithmetic ended up being nearly identical to how you solved it by looking for wraparounds in modular. I started with 3=8=13=18… (mod 5) and then I did 14=26=66=92=118… (mod 26) which is pretty much what you did. So yes, not only is it a fairly elementary application for someone who’s studied it, it’s a fairly elementary application for someone who hasn’t!

What’s fun about this is from the pattern we can extend to any other with relative ease, lets say we want to find the first yellow H:

5=10=15=20…. mod(5)

8=34=50…. mod(26)

So N=50.

It really is the same thing as the first method but I suppose if nothing else it’s expressed in more concrete notation.