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

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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>).
  
==Video Solution==
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==Solution 3 (Variant of Solution 1, but using Brute Force)==
https://youtu.be/m4a5kIRdnOM - Happytwin
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After finding out that the last digit must be <math>5</math>, the number is of the form <math>5\square 5\square 5</math>. If the unknown digit is <math>x</math>, we can find that one of the solutions to <math>x</math> is <math>0</math>, since <math>5+5+5</math> is equal to <math>15</math>, which is divisible by <math>3</math>. After trying every one digit number, you'll notice that <math>x</math> must be a multiple of <math>3</math>, meaning that <math>x=0</math>, <math>3</math>, <math>6</math>, or <math>9</math>. <math>50505</math>, <math>53535</math>, <math>56565</math>, and <math>59595</math> are the <math>\boxed{\textbf{(B) }4}</math> solutions to this question.
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==Video Solution by North America Math Contest Go Go Go==
  
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https://www.youtube.com/watch?v=5Qo4pG3Uk_U
  
https://youtu.be/SPNobOd4t1c (Channel also has resources to prepare for your AIME qualification)
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~North America Math Contest Go Go Go
  
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==Video Solution by WhyMath==
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https://youtu.be/8nVSeTx5rro
  
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~savannahsolver
  
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==Video Solution==
 
https://youtu.be/VnOecUiP-SA
 
https://youtu.be/VnOecUiP-SA
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==Video Solution by Interstigation==
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https://youtu.be/YnwkBZTv5Fw?t=980
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~Interstigation
  
 
==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 15:16, 4 January 2022

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 (Variant of Solution 1, but using Brute Force)

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.

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://youtu.be/VnOecUiP-SA

Video Solution by Interstigation

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

~Interstigation

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
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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