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

(Created page with "==Problem 19== A number is called flippy if its digits alternate between two distinct digits. For example, <math>2020</math> and <math>37373</math> are flippy, but <math>3883<...")
 
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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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==See also==
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{{AMC8 box|year=2020|num-b=18|num-a=20}}
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{{MAA Notice}}

Revision as of 00:20, 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

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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