2020 AMC 8 Problems/Problem 19
Contents
- 1 Problem 19
- 2 Solution 1
- 3 Solution 2 (variant of Solution 1)
- 4 Solution 3 (Variant of Solution 1, but using Brute Force)
- 5 Solution 4 (mods)
- 6 Video Solution by OmegaLearn
- 7 Video Solution by North America Math Contest Go Go Go
- 8 Video Solution by WhyMath
- 9 Video Solution
- 10 Video Solution by Interstigation
- 11 Video Solution (CLEVER MANIPULATIONS!!!)
- 12 See also
Problem 19
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
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 (variant of Solution 1)
As in Solution 1, we find that such numbers must start with and alternate with (i.e. must be of the form ), where the two digits between the s need to be the same. Call that digit . For the number to be divisible by , the sum of the digits must be divisible by ; since the sum of the three s is , which is already a multiple of , it must also be the case that is a multiple of . Thus, the problem reduces to finding the number of digits from to for which is a multiple of . This leads to , , , or , so there are possible numbers (namely , , , and ).
Solution 3 (Variant of Solution 1, but using Brute Force)
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.
Solution 4 (mods)
assume the number is Solutions:
Video Solution by OmegaLearn
https://youtu.be/At4w8uylvv8?t=692
~ pi_is_3.14
Video Solution by North America Math Contest Go Go Go
https://www.youtube.com/watch?v=5Qo4pG3Uk_U
~North America Math Contest Go Go Go
Video Solution by WhyMath
~savannahsolver
Video Solution
https://www.youtube.com/watch?v=a3Z7zEc7AXQ
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=980
~Interstigation
Video Solution (CLEVER MANIPULATIONS!!!)
~Education, the Study of Everything
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
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