Difference between revisions of "2020 AMC 10A Problems/Problem 6"

(Solution)
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The units digit, for all numbers divisible by 5, must be either <math>0</math> or <math>5</math>. However, since all digits are even, the units digit must be <math>0</math>. The middle two digits can be 0, 2, 4, 6, or 8, giving 5 choices for each. The first (thousands) digit can be 2, 4, 6, or 8, giving 4 choices. Therefore we get <math>4\times 5 \times 5 \times 1 = \boxed{\textbf{(B) } 100}</math> possible 4-digit integers.
 
The units digit, for all numbers divisible by 5, must be either <math>0</math> or <math>5</math>. However, since all digits are even, the units digit must be <math>0</math>. The middle two digits can be 0, 2, 4, 6, or 8, giving 5 choices for each. The first (thousands) digit can be 2, 4, 6, or 8, giving 4 choices. Therefore we get <math>4\times 5 \times 5 \times 1 = \boxed{\textbf{(B) } 100}</math> possible 4-digit integers.
  
==Video Solution==
+
==Video Solution 1==
  
 
Education, the Study of Everything
 
Education, the Study of Everything
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https://youtu.be/pvqpXWAvtAk
 
https://youtu.be/pvqpXWAvtAk
  
 +
==Video Solution 2==
 
https://youtu.be/JEjib74EmiY
 
https://youtu.be/JEjib74EmiY
  
 
~IceMatrix
 
~IceMatrix
  
 +
==Video Solution 3==
 
https://youtu.be/Ep6XF3VUO3E
 
https://youtu.be/Ep6XF3VUO3E
  

Revision as of 14:33, 22 December 2020

The following problem is from both the 2020 AMC 12A #4 and 2020 AMC 10A #6, so both problems redirect to this page.

Problem

How many $4$-digit positive integers (that is, integers between $1000$ and $9999$, inclusive) having only even digits are divisible by $5?$

$\textbf{(A) } 80 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 125 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 500$

Solution

The units digit, for all numbers divisible by 5, must be either $0$ or $5$. However, since all digits are even, the units digit must be $0$. The middle two digits can be 0, 2, 4, 6, or 8, giving 5 choices for each. The first (thousands) digit can be 2, 4, 6, or 8, giving 4 choices. Therefore we get $4\times 5 \times 5 \times 1 = \boxed{\textbf{(B) } 100}$ possible 4-digit integers.

Video Solution 1

Education, the Study of Everything

https://youtu.be/pvqpXWAvtAk

Video Solution 2

https://youtu.be/JEjib74EmiY

~IceMatrix

Video Solution 3

https://youtu.be/Ep6XF3VUO3E

~savannahsolver

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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 AMC 10 Problems and Solutions
2020 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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 AMC 12 Problems and Solutions

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

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