Difference between revisions of "2020 AMC 10A Problems/Problem 6"
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== 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, giving 5 choices for each. The first (thousands) digit can be 2, 4, 6, or 8, giving 4 choices. Therefore | + | 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, giving 5 choices for each. The first (thousands) digit can be 2, 4, 6, or 8, giving 4 choices. Therefore we get <math>4\times 5 \times 5 \times 1 = \boxed{\textbf{(B) } 100}</math> possible 4-digit integers. |
==Video Solution== | ==Video Solution== | ||
Revision as of 23:44, 28 November 2020
- The following problem is from both the 2020 AMC 12A #4 and 2020 AMC 10A #6, so both problems redirect to this page.
Contents
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, giving 5 choices for each. The first (thousands) digit can be 2, 4, 6, or 8, giving 4 choices. Therefore we get 
possible 4-digit integers.
Video Solution
Education, the Study of Everything
~IceMatrix
~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. 
