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> | <math>\textbf{(A) } 80 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 125 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 500</math> | ||
| − | == Solution == | + | == 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, we have 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== | ==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}} | ||
| + | {{AMC12 box|year=2020|ab=A|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
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 
-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, we have 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. 
