Difference between revisions of "2020 AMC 8 Problems/Problem 19"
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− | To be divisible by <math>15</math>, a number must first be divisible by <math>3</math> and <math>5</math>. By divisibility rules, the last digit must be either <math>5</math> or <math>0</math>, and the sum of the digits must be divisible by <math>3</math>. If the last digit is <math>0</math>, the first digit would be <math>0</math> (because the digits alternate). So, the last digit must be <math>5</math>, and we have <cmath>5+x+5+x+5 \equiv 0 \pmod{3}</cmath> <cmath>2x \equiv 0 \pmod{3}</cmath> We know the inverse exists because 2 is relatively prime to 3, and thus we can conclude that <math>x</math> (or the second and fourth digits) is always a multiple of <math>3</math>. We have 4 options: <math>0, 3, 6, 9</math>, and our answer is <math>4</math> and <math>\boxed{B}</math> ~samrocksnature | + | To be divisible by <math>15</math>, a number must first be divisible by <math>3</math> and <math>5</math>. By divisibility rules, the last digit must be either <math>5</math> or <math>0</math>, and the sum of the digits must be divisible by <math>3</math>. If the last digit is <math>0</math>, the first digit would be <math>0</math> (because the digits alternate). So, the last digit must be <math>5</math>, and we have <cmath>5+x+5+x+5 \equiv 0 \pmod{3}</cmath> <cmath>2x \equiv 0 \pmod{3}</cmath> We know the inverse exists because 2 is relatively prime to 3, and thus we can conclude that <math>x</math> (or the second and fourth digits) is always a multiple of <math>3</math>. We have 4 options: <math>0, 3, 6, 9</math>, and our answer is <math>4</math> and <math>\boxed{\textbf{(B) } 4}</math> ~samrocksnature |
==Solution 2== | ==Solution 2== |
Revision as of 11:56, 18 November 2020
Contents
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
To be divisible by , a number must first be divisible by and . By divisibility rules, the last digit must be either or , and the sum of the digits must be divisible by . If the last digit is , the first digit would be (because the digits alternate). So, the last digit must be , and we have We know the inverse exists because 2 is relatively prime to 3, and thus we can conclude that (or the second and fourth digits) is always a multiple of . We have 4 options: , and our answer is and ~samrocksnature
Solution 2
A number that is divisible by must be divisible by and . To be divisible by , the sum of its digits must be divisible by and to be divisible by , it must end in a or a . Observe that a five-digit flippy number must start and end with the same digit. Since a five-digit number cannot start with , it also cannot end in . This means that the numbers that we are looking for must end in . This also means that they must start with and alternate with (i.e. the number must be of the form ). The two digits between the must be the same. Let's call that digit . We know that the sum of the digits must be a multiple of . Since the sum of the three is which is already a multiple of , for the entire five-digit number to be a multiple of , it must also be the case that is also a multiple of . Thus, the problem reduces to finding the number of digits from to for which is a multiple of . This leads to and and we have four valid answers (i.e. and ) .
~ junaidmansuri
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 | ||
All AJHSME/AMC 8 Problems and Solutions |
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