Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (Minor improvement in wording)
(→‎Video Solution by Math-X (First understand the problem!!!))
 
(23 intermediate revisions by 16 users not shown)
Line 1: Line 1:
−
==Problem==
+
==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</math> and <math>123123</math> are not. How many five-digit flippy numbers are divisible by <math>15?</math>
 
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</math> and <math>123123</math> are not. How many five-digit flippy numbers are divisible by <math>15?</math>
  
Line 9: Line 9:
 
==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>).
 +
 +
==Solution 3==
 +
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.
 +
 +
==Solution 4 (mods)==
 +
assume the number is <math>ababa</math>
 +
<math>10101a+1010b=0 (mod 15)\newline</math>
 +
<math>6a+5b=0 (mod 15)\newline</math>
 +
<math>a=0 (mod 5)\newline</math>
 +
<math>5b=0 (mod 15)\newline</math>
 +
<math>b=0 (mod 3)\newline</math>
 +
Solutions: <math>(5,0),(5,3),(5,6),(5,9)\newline</math>
 +
<math>\boxed{4}</math>
 +
 +
==Video Solution by NiuniuMaths (Easy to understand!)==
 +
https://www.youtube.com/watch?v=bHNrBwwUCMI
 +
 +
~NiuniuMaths
 +
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/UnVo6jZ3Wnk?si=h40WB_9lDM1MmNAF&t=3548
 +
 +
~Math-X
 +
 +
==Video Solution (🚀Very Fast🚀)==
 +
https://youtu.be/4Mvm5u4RT6E
 +
 +
~Education, the Study of Everything
 +
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/At4w8uylvv8?t=692
 +
 +
~ pi_is_3.14
 +
 +
==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://www.youtube.com/watch?v=a3Z7zEc7AXQ
 +
 +
==Video Solution by Interstigation==
 +
https://youtu.be/YnwkBZTv5Fw?t=980
 +
 +
~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}}

Latest revision as of 15:35, 26 January 2024

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

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.

Solution 4 (mods)

assume the number is $ababa$ $10101a+1010b=0 (mod 15)\newline$ $6a+5b=0 (mod 15)\newline$ $a=0 (mod 5)\newline$ $5b=0 (mod 15)\newline$ $b=0 (mod 3)\newline$ Solutions: $(5,0),(5,3),(5,6),(5,9)\newline$ $\boxed{4}$

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=bHNrBwwUCMI

~NiuniuMaths

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/UnVo6jZ3Wnk?si=h40WB_9lDM1MmNAF&t=3548

~Math-X

Video Solution (🚀Very Fast🚀)

https://youtu.be/4Mvm5u4RT6E

~Education, the Study of Everything

Video Solution by OmegaLearn

https://youtu.be/At4w8uylvv8?t=692

~ pi_is_3.14

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://www.youtube.com/watch?v=a3Z7zEc7AXQ

Video Solution by Interstigation

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

~Interstigation

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_19&oldid=213193"