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Difference between revisions of "2020 AMC 8 Problems/Problem 19"

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

Revision as of 10:06, 20 November 2020

Problem

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 $59595$).

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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