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2020 AMC 8 Problems/Problem 19

Problem 19

A number is called flippy if its digits alternate between two distinct digits. For example, $2020$ and $37373$ are flippy, but $3883$ and $123123$ are not. How many five-digit flippy numbers are divisible by $15?$

$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8$

Solution 1

A number is divisible by $15$ precisely if it is divisible by $3$ and $5$. The latter means the last digit must be either $5$ or $0$, and the former means the sum of the digits must be divisible by $3$. If the last digit is $0$, the first digit would be $0$ (because the digits alternate), which is not possible. Hence the last digit must be $5$, and the number is of the form $5\square 5\square 5$. If the unknown digit is $x$, we deduce $5+x+5+x+5 \equiv 0 \pmod{3} \Rightarrow 2x \equiv 0 \pmod{3}$. We know $2^{-1}$ exists modulo $3$ because 2 is relatively prime to 3, so we conclude that $x$ (i.e. the second and fourth digit of the number) must be a multiple of $3$. It can be $0$, $3$, $6$, or $9$, so there are $\boxed{\textbf{(B) }4}$ options: $50505$, $53535$, $56565$, and $59595$.

Solution 2 (variant of Solution 1)

As in Solution 1, we find that such numbers must start with $5$ and alternate with $5$ (i.e. must be of the form $5\square 5\square 5$), where the two digits between the $5$s need to be the same. Call that digit $x$. For the number to be divisible by $3$, the sum of the digits must be divisible by $3$; since the sum of the three $5$s is $15$, which is already a multiple of $3$, it must also be the case that $x+x=2x$ is a multiple of $3$. Thus, the problem reduces to finding the number of digits from $0$ to $9$ for which $2x$ is a multiple of $3$. This leads to $x=0$, $3$, $6$, or $9$, so there are $\boxed{\textbf{(B) }4}$ possible numbers (namely $50505$, $53535$, $56565$, and $59595$).

Solution 3

After finding out that the last digit must be $5$, the number is of the form $5\square 5\square 5$. If the unknown digit is $x$, we can find that one of the solutions to $x$ is $0$, since $5+5+5$ is equal to $15$, which is divisible by $3$. After trying every one digit number, you'll notice that $x$ must be a multiple of $3$, meaning that $x=0$, $3$, $6$, or $9$. $50505$, $53535$, $56565$, and $59595$ are the $\boxed{\textbf{(B) }4}$ solutions to this question.

Solution 4 (mods)

assume the number is $ababa$ $10101a+1010b=0 (mod 15)\newline$ $6a+5b=0 (mod 15)\newline$ $a=0 (mod 5)\newline$ $5b=0 (mod 15)\newline$ $b=0 (mod 3)\newline$ Solutions: $(5,0),(5,3),(5,6),(5,9)\newline$ $\boxed{4}$

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=bHNrBwwUCMI

~NiuniuMaths

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/UnVo6jZ3Wnk?si=h40WB_9lDM1MmNAF&t=3548

~Math-X

Video Solution (🚀Very Fast🚀)

https://youtu.be/4Mvm5u4RT6E

~Education, the Study of Everything

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

https://youtu.be/8nVSeTx5rro

~savannahsolver

Video Solution

https://www.youtube.com/watch?v=a3Z7zEc7AXQ

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=980

~Interstigation

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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