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

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==Solution 2 (variant of Solution 1)==
 
==Solution 2 (variant of Solution 1)==
 
As in Solution 1, we find that such numbers must start with <math>5</math> and alternate with <math>5</math> (i.e. must be of the form <math>5\square 5\square 5</math>), where the two digits between the <math>5</math>s need to be the same. Call that digit <math>x</math>. For the number to be divisible by <math>3</math>, the sum of the digits must be divisible by <math>3</math>; since the sum of the three <math>5</math>s is <math>15</math>, which is already 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</math>, <math>3</math>, <math>6</math>, or <math>9</math>, so there are <math>\boxed{\textbf{(B) }4}</math> possible numbers (namely <math>50505</math>, <math>53535</math>, <math>56565</math>, and <math>59595</math>).
 
As in Solution 1, we find that such numbers must start with <math>5</math> and alternate with <math>5</math> (i.e. must be of the form <math>5\square 5\square 5</math>), where the two digits between the <math>5</math>s need to be the same. Call that digit <math>x</math>. For the number to be divisible by <math>3</math>, the sum of the digits must be divisible by <math>3</math>; since the sum of the three <math>5</math>s is <math>15</math>, which is already 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</math>, <math>3</math>, <math>6</math>, or <math>9</math>, so there are <math>\boxed{\textbf{(B) }4}</math> possible numbers (namely <math>50505</math>, <math>53535</math>, <math>56565</math>, and <math>59595</math>).
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 +
==Video Solution==
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https://youtu.be/8nVSeTx5rro
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 +
~savannahsolver
  
 
==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 17:34, 23 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 $59595$).

Video Solution

https://youtu.be/8nVSeTx5rro

~savannahsolver

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AJHSME/AMC 8 Problems and Solutions

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