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

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{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #4]] and [[2020 AMC 10A Problems|2020 AMC 10A #6]]}}
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==Problem==
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How many <math>4</math>-digit positive integers (that is, integers between <math>1000</math> and <math>9999</math>, inclusive) having only even digits are divisible by <math>5?</math>
 
How many <math>4</math>-digit positive integers (that is, integers between <math>1000</math> and <math>9999</math>, inclusive) having only even digits are divisible by <math>5?</math>
  
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== Solution ==
 
== Solution ==
First, we need to note that the digits have to be even, but since only one even digit for the units digit (<math>0</math>) we get <math>4\cdot5\cdot5\cdot1=\boxed{(B)100}</math>.
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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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swrebby
  
-middletonkids
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==Video Solution==
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https://youtu.be/JEjib74EmiY
  
== Solution 2 ==
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~IceMatrix
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 (0) 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
 
  
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2020|ab=A|num-b=5|num-a=7}}
 
{{AMC10 box|year=2020|ab=A|num-b=5|num-a=7}}
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{{AMC12 box|year=2020|ab=A|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:55, 18 March 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 ones digit, for all numbers divisible by 5, must be either $0$ or $5$. However, from the restriction in the problem, it must be even, giving us exactly one choice ($0$) for this digit. For the middle two digits, we may choose any even integer from $[0, 8]$, meaning that we have $5$ total options. For the first digit, we follow similar intuition but realize that it cannot be $0$, hence giving us 4 possibilities. Therefore, using the multiplication rule, we get $4\times 5 \times 5 \times 1 = \boxed{\textbf{(B) } 100}$. ~ciceronii swrebby

Video Solution

https://youtu.be/JEjib74EmiY

~IceMatrix

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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