Logical Strategy: Casework

The Census Taker Problem

A census taker approaches a house and asks the woman who answers the door "How many children do you have, and what are their ages?"

The woman replies "I have three children, the product of their ages are 36, the sum of their ages are equal to the address of the house next door."

The census taker walks next door, comes back and says "I need more information."

The woman replies "I have to go, my oldest child is sleeping upstairs."

Census taker then says "Thank you, I now have everything I need."

At first sight, this problem seems inpenetratable. How in the world are the second and third reply of the woman going to help us? It turns out that they're the key of determining the answer.

The product of the ages is $36$, so there are few possibilities :

$1, 1, 36$ Sum : $38$

$1, 2, 18$ Sum : $21$

$1, 3, 12$ Sum : $16$

$1, 4, 9$ Sum : $14$

$1, 6, 6$ Sum : $13$

$2, 2, 9$ Sum : $13$

$2, 3, 6$ Sum : $11$

$3, 3, 4$ Sum : $10$

Try to solve the problem from here. Click Continue when you're ready for the next part of the solution.

Now we know what the second reply means! Since the census taker still can't determine their ages after knowing the sum, it follows that the sum must be $13$, so that there could be ambiguity. The last statement states that the oldest child exist, so the ages must be $2, 2, 9$!

Suppose a task can be done by either method one or method two. If there are $m$ ways to use method one, $n$ ways to use method two and there is no way common to methods one and two, then there are $m + n$ ways to do the task.

In general, you may need to consider more than two cases. The most important thing you can do is stay organized.

Example 1: How many bit strings are there with length less than 5?

Solution: $2 + 2^2+2^3 + 2^4 = \boxed{30.}$

Example 2: If a password must be six to eight characters long, each character is either an uppercase letter or a digit, and each password must contain at least one digit. How many possible passwords are there?

Solution: Let $P_6, P_7,$ and $P_8$ denote the number of passwords of length 6, 7 and 8, respectively. We know that $P = P_6 + P_7 + P_8.$ From complementary counting, we know that $$P_6 = 36^6 - 26^6, P_7= 36^7 - 26^7, P_8 = 36^8 - 26^8.$$ Our answer is the sum of all of these.

The only way to get better at casework is to do more problems!