2020 AMC 10A Problems/Problem 6

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 1

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

Solution 2

The ones digit, for all the numbers that have to divisible be 5, must be a $0$ or a $5$ . Since the problem states that we can only use even digits, the last digit must be $0$ . From there, there are no other restrictions since the divisibility rule for 5 states that the last digit must be a $0$ or a $5$ . So there are $4$ even digit options for the first number then $5$ for the middle 2. So when we have to do $4 \cdot 5 \cdot 5 \cdot 1 = \boxed{\textbf{(B) } 100}$ . ~bobthefam

Solution 3

This is basically 4*5*5 which is 100

~math31415926535

Video Solution

https://youtu.be/JEjib74EmiY

~IceMatrix

https://youtu.be/Ep6XF3VUO3E

~savannahsolver

2020 AMC 10A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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.

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