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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{B}</math>    ~samrocksnature
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==Solution 2==
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A number that is divisible by <math>15</math> must be divisible by <math>3</math> and <math>5</math>. To be divisible by <math>3</math>, the sum of its digits must be divisible by <math>3</math> and to be divisible by <math>5</math>, it must end in a <math>5</math> or a <math>0</math>. Observe that a five-digit flippy number must start and end with the same digit. Since a five-digit number cannot start with <math>0</math>, it also cannot end in <math>0</math>. This means that the numbers that we are looking for must end in <math>5</math>. This also means that they must start with <math>5</math> and alternate with <math>5</math> (i.e. the number must be of the form <math>5\square5\square5</math>). The two digits between the <math>5s</math> must be the same. Let's call that digit <math>x</math>. We know that the sum of the digits must be a multiple of <math>3</math>. Since the sum of the three <math>5s</math> is <math>15</math> which is already a multiple of <math>3</math>, for the entire five-digit number to be a multiple of <math>3</math>, it must also be the case that <math>x+x=2x</math> is a multiple of <math>3</math>. Thus, the problem reduces to finding the number of digits from <math>0</math> to <math>9</math> for which <math>2x</math> is a multiple of <math>3</math>. This leads to <math>x=0,3,6,</math> and <math>9</math> and we have four valid answers (i.e. <math>50505,53535,56565,</math> and <math>59595</math>) <math>\implies\boxed{\textbf{(B) } 4}</math>.<br>
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~ junaidmansuri
  
 
==See also==  
 
==See also==  
 
{{AMC8 box|year=2020|num-b=18|num-a=20}}
 
{{AMC8 box|year=2020|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 06:42, 18 November 2020

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

To be divisible by $15$, a number must first be divisible by $3$ and $5$. By divisibility rules, the last digit must be either $5$ or $0$, and 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). So, the last digit must be $5$, and we have \[5+x+5+x+5 \equiv 0 \pmod{3}\] \[2x \equiv 0 \pmod{3}\] We know the inverse exists because 2 is relatively prime to 3, and thus we can conclude that $x$ (or the second and fourth digits) is always a multiple of $3$. We have 4 options: $0, 3, 6, 9$, and our answer is $4$ and $\boxed{B}$ ~samrocksnature

Solution 2

A number that is divisible by $15$ must be divisible by $3$ and $5$. To be divisible by $3$, the sum of its digits must be divisible by $3$ and to be divisible by $5$, it must end in a $5$ or a $0$. Observe that a five-digit flippy number must start and end with the same digit. Since a five-digit number cannot start with $0$, it also cannot end in $0$. This means that the numbers that we are looking for must end in $5$. This also means that they must start with $5$ and alternate with $5$ (i.e. the number must be of the form $5\square5\square5$). The two digits between the $5s$ must be the same. Let's call that digit $x$. We know that the sum of the digits must be a multiple of $3$. Since the sum of the three $5s$ is $15$ which is already a multiple of $3$, for the entire five-digit number to be 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 $x=0,3,6,$ and $9$ and we have four valid answers (i.e. $50505,53535,56565,$ and $59595$) $\implies\boxed{\textbf{(B) } 4}$.
~ 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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