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

(deleted first solution, as second solution used same method but elaborated more)
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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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{{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 08:02, 1 February 2020

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

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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