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]]}} | ||
+ | |||
+ | ==Problem== | ||
+ | |||
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> | ||
+ | |||
+ | <math>\textbf{(A) } 80 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 125 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 500</math> | ||
+ | |||
+ | == Solution== | ||
+ | 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, so there is a total of <math>4\cdot5\cdot5\cdot1 = \boxed{\textbf{(B) } 100} \qquad</math> 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 | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC10 box|year=2020|ab=A|num-b=5|num-a=7}} | ||
+ | {{AMC12 box|year=2020|ab=A|num-b=3|num-a=5}} | ||
+ | {{MAA Notice}} |
Latest revision as of 09:21, 7 October 2022
- 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 -digit positive integers (that is, integers between and , inclusive) having only even digits are divisible by
Solution
The units digit, for all numbers divisible by 5, must be either or . However, since all digits are even, the units digit must be . 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, so there is a total of numbers.
Video Solution 1
Education, the Study of Everything
Video Solution 2
~IceMatrix
Video Solution 3
~savannahsolver
See Also
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.