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How
to Write a Solution
Clear Casework
by
Richard Rusczyk
Sometimes the solution to a problem comes down to investigating a few different cases. In your solution, you should identify the cases clearly and show that these cases cover all possibilities.
Here's a sample problem:
(This problem comes from the Art of Problem Solving MATHCOUNTS/AMC Counting course.)
The solution above is short, and the answer is correct, but it's not at all clear that all possibilities have been discovered. Also, it's pretty tough to see that we have found exactly 52 solutions - the reader is forced to go through and count herself.
The solution below clearly covers all possible cases and leaves no doubt that the total is 52.
Case 1: The smallest digit is 0.
If the smallest digit is 0, then the number must contain a second 0. Thus, this case consists of numbers of the form n00, where 1 <= n <= 9 is any digit from 1 to 9. There are thus 9 numbers with smallest digit 0 that satisfy the problem.
Case 2: The smallest digit is 1.
If the smallest digit is 1, the number must be of the form nn1, or permutations of this form (i.e. n1n or 1nn). However, these 3 permutations are the same when n = 1. Hence, we have 3 permutations each for 2 <= n <= 9 and only 1 for n = 1, for a total of 1 + 3(8) = 25 numbers with smallest digit 1 that satisfy the problem.
Case 3: The smallest digit is 2.
If the smallest digit is 2, then the number is of the form 2mn, where n =2m, and permutations of this form. Our only options here are (m,n) = (2,4), which gives us 3 numbers (224, 242, 422), (m,n) = (3,6), which gives us 6 numbers (permutations of 236), and (m,n) = (4,8), which also gives us 6 numbers. Hence, there are 3 + 6 + 6 = 15 numbers with smallest digit 2.
Case 4: The smallest digit is 3.
There are 3 solutions in this case: 339, 393, 933.
Case 5: The smallest digit is larger than 3.
If the smallest digit is larger than 3, the smallest product we can form with two of the digits is 4(4) = 16, which is not a single digit number. Hence, there are no numbers that satisfy the problem with smallest digit larger than 3.
Since every possible 3-digit number falls in exactly one of these cases, we conclude that there are
9 + 25 + 15 + 3 = 52
numbers that satisfy the problem.
