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

Latest revision as of 03:01, 30 March 2024

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, but the thousands digit can only be 2, 4, 6, or 8 since it cannot be zero. There is one choice for the units digit, 5 choices for each of the middle 2 digits, and 4 choices for the thousands digit, we have a total of $4\cdot5\cdot5\cdot1 = \boxed{\textbf{(B) } 100} \qquad$ numbers.

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

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.

@@ Line 7: / Line 7: @@
 <math>\textbf{(A) } 80 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 125 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 500</math>
-== Solution 1==
+== 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
+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, but the thousands digit can only be 2, 4, 6, or 8 since it cannot be zero. There is one choice for the units digit, 5 choices for each of the middle 2 digits, and 4 choices for the thousands digit, we have a total of <math>4\cdot5\cdot5\cdot1 = \boxed{\textbf{(B) } 100} \qquad</math> numbers.
-swrebby
-== Solution 2==
+==Video Solution 1==
-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
-== Solution 3==
-This is basically 4*5*5 which is 100
- ~math31415926535
-==Video Solution==
+Education, the Study of Everything
+https://youtu.be/pvqpXWAvtAk
+==Video Solution 2==
 https://youtu.be/JEjib74EmiY
 ~IceMatrix
+==Video Solution 3==
 https://youtu.be/Ep6XF3VUO3E

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