Problem of the Week
Welcome to the Centenary Math Problem of the Week contest! You can always find the most recent problem of the week below. You can also look at the current scores or read the rules.
Fall 2008 Problem 4: Is that enough information?
Mathematicians Alice and Bob were having a discussion.
Alice: Did you know I have four daughters?
Bob: No, I didn't. What are their ages?
Alice: Well, the product of their ages is 72 and the sum of their ages equals the age of your oldest daughter.
Bob (thinking for a minute): No, that's not enough information for me to figure out their ages. Can you tell me something more?
Alice: My oldest daughter carries around my youngest daughter all the time.
Bob: Oh great —- now I know their ages!
What are the ages of the four daughters? This is a three point problem. I will accept answers through noon on Wednesday 10/8. I will give bonus points for the first correct answer and the best explanation.
Fall 2008 Problem 3: C-E-N-T-E-N-A-R-Y
You want to spell Centenary using the picture below. You start at the letter C and then move either straight down to the E below or to the E below and one letter to the right. At every letter you repeat this process: you either move straight down or move down and to the right. (So, for example, from the second N on the third row, you can move to the second T on the fourth row or the third T on the fourth row.) You never move up or to the left.
Following this procedure, in how many different ways can you spell CENTENARY?
Solution
Our first correct solution came from Matthew Chumley, earning him three points. Brent Krise also had the best explanation (see below), earning him 3 points.
There are 256 different ways to spell CENTENARY in this fashion. To figure this problem out, I started with a smaller word. I used a three letter word and found 4 different ways to spell it. Then I moved to a four letter word and found there were 8 different ways to spell it. It seemed that a pattern was occurring where the number of different ways, P(n) was equal to 2^(n-1), where n was the length of the word. To assure my assumption was correct I proved it by induction. The base case when n=1 is simple, there is only 1 way. Now for the inductive step, assume P(n)=2^(n-1) and show P(n+1) = 2^(n+1-1) = 2^(n). We know that P(n+1) can be rewritten as P(n)*2 since we are just doubling the ways from n to n+1. Then subbing P(n)=2^(n-1) we get 2*(n-1)*2 = 2^n. Thus the number of ways to spell CENTENARY is 2^(9-1) = 256.
Other correct solutions (earning three points) came from Sterling Williams and Michaela Berg. Correct solutions from non-Centenary students came from Mark Goadrich, Scott Jackson, and Chris Evert.