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

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<math>\textbf{(A) } 80 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 125 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 500</math>
 
<math>\textbf{(A) } 80 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 125 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 500</math>
  
== Solution 1==
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== Solution==
The ones digit, for all numbers divisible by 5, must be either <math>0</math> or <math>5</math>. However, from the restriction in the problem, it must be even, giving us exactly one choice (<math>0</math>) for this digit. For the middle two digits, we may choose any even integer from <math>[0, 8]</math>, meaning that we have <math>5</math> total options. For the first digit, we follow similar intuition but realize that it cannot be <math>0</math>, hence giving us 4 possibilities. Therefore, using the multiplication rule, we get <math>4\times 5 \times 5 \times 1 = \boxed{\textbf{(B) } 100}</math>. ~ciceronii
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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.
swrebby
 
  
== Solution 2==
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==Video Solution==
The ones digit, for all the numbers that have to divisible be 5, must be a <math>0</math> or a <math>5</math>. Since the problem states that we can only use even digits, the last digit must be <math>0</math>. From there, there are no other restrictions since the divisibility rule for 5 states that the last digit must be a <math>0</math> or a <math>5</math>. So there are <math>4</math> even digit options for the first number then <math>5</math> for the middle 2. So when we have to do <math>4 \cdot 5 \cdot 5 \cdot 1 = \boxed{\textbf{(B) } 100}</math>. ~bobthefam
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== Solution 3==
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Education, the Study of Everything
This is basically 4*5*5 which is 100
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https://youtu.be/pvqpXWAvtAk
  
==Video Solution==
 
 
https://youtu.be/JEjib74EmiY
 
https://youtu.be/JEjib74EmiY
  

Revision as of 00:44, 29 November 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

Education, the Study of Everything

https://youtu.be/pvqpXWAvtAk

https://youtu.be/JEjib74EmiY

~IceMatrix

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

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