2020 AMC 8
Complete problem set with solutions and individual problem pages
A number is called flippy if its digits alternate between two distinct digits. For example, and are flippy, but and are not. How many five-digit flippy numbers are divisible by
- A.
- B.
- C.
- D.
- E.
Solution 1
A number is divisible by precisely if it is divisible by and . The latter means the last digit must be either or , and the former means the sum of the digits must be divisible by . If the last digit is , the first digit would be (because the digits alternate), which is not possible. Hence the last digit must be , and the number is of the form . If the unknown digit is , we deduce . We know exists modulo because 2 is relatively prime to 3, so we conclude that (i.e. the second and fourth digit of the number) must be a multiple of . It can be , , , or , so there are options: , , , and .
Solution 2
After finding out that the last digit must be , the number is of the form . If the unknown digit is , we can find that one of the solutions to is , since is equal to , which is divisible by . After trying every one digit number, you'll notice that must be a multiple of , meaning that , , , or . , , , and are the solutions to this question.
