Universe

Can You Solve the Crazy Cat Lady Puzzle?

February 5, 2016 | Elizabeth Knowles

If you chose a pair of Claire's cats at random, the probability that they would both have white tails would be exactly 1/2. Claire has fewer than 20 cats. How many does she have?

Or are you just as confused as Tom?

When asked about how many cats they have, most people would reply with a number. Claire, on the other hand, is a little unconventional, so she prefers to reply with a riddle. Tom doesn’t think that he has enough information to figure it out, but maybe you can help him.

Let’s see what we know:

-       Claire has fewer than 20 cats.

-       The number of distinct pairs of cats with white tails is half of the number of possible pairs.

We can calculate the number of possible pairs as c(c-1)/2, where c denotes the total number of cats. If that doesn’t make sense to you, think of it as how many ways you can pick a pair of socks from, say, three socks (a red, a blue and a green one).

Red and Blue
Red and Green
Blue and Green
Blue and Red
Green and Red

Green and Blue

You have three options for the first sock, and two options for the second sock, but you have to divide the total by two because the pair “red and blue” is the same as the pair “blue and red.” You may look a little funny wearing two different colored socks though!

By the same logic, you know that there are w(w-1)/2 pairs of white-tailed cats if w denotes the total number of white-tailed cats.

So, based on what we know we can say that (w(w-1)/2)/(c(c-1)/2)=0.5

Since we’re dividing both the top and bottom of the left side of the equation by 2, we can cancel it out so we get:

w(w-1)/c(c-1)=0.5
or

w(w-1)=0.5c(c-1)

You’ve probably heard before that you need two equations to solve for two variables, but we have an extra piece of information that we can use: Claire has fewer than 20 cats.

She obviously can’t have just one cat because she’s talking about pairs.

SEE ALSO: Can You Solve the Coin Puzzle?

She can’t have just two cats because then the pair would be either white-tailed or not white-tailed. The fraction wouldn’t be half.

If she had 4 cats, we’d get:

w(w-1)=6

w=3

That works! Let’s see if there are any other possibilities…

The right hand side of the equation must lead to a number that we can factor into two numbers that are just one apart.

Number of Cats (c)

Right-hand side

Factor pairs

5
10
2 and 5
6
15
3 and 5
7
21
3 and 7
8
28
2 and 14
4 and 7
9
36
2 and 18
3 and 12
4 and 9
6 and 6
10
45
3 and 15
5 and 9
11
55
5 and 11
12
66
2 and 33
3 and 22
6 and 11
13
78
2 and 39
3 and 26
6 and 13
14
91
7 and 13
15
105
3 and 35
5 and 21
7 and 15
16
120
2 and 60
3 and 40
4 and 30
5 and 24
6 and 20
8 and 15
10 and 12
17
136
2 and 68
4 and 34
8 and 17
18
153
3 and 51
9 and 17
19
171
3 and 57
9 and 19

None of the factor pairs here are just one apart and since we know that she has fewer than 20 cats, those are all of the numbers we have to test for!

If Tom didn’t know that she had fewer than 20 cats, then he might have thought she had 21 (with 15 white ones), but we don’t have to consider that case (or any larger ones).

So, Claire has 4 cats and 3 of them have white tails.

This puzzle was adapted from one from “Professor Stewart’s Cabinet of Mathematical Curiosities” by Ian Stewart.

Try more brain teasers here.

Hot Topics

Facebook comments