2020 AMC 10A Problems/Problem 6: Difference between revisions
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== Solution == | == Solution == | ||
First, we need to note that the digits have to be even, but since only one even digit for the units digit (<math>0</math>) we get <math>4\cdot5\cdot5\cdot1=\boxed{(B)100}</math>. | First, we need to note that the digits have to be even, but since only one even digit for the units digit (<math>0</math>) we get <math>4\cdot5\cdot5\cdot1=\boxed{(B)100}</math>. | ||
-middletonkids | -middletonkids | ||
Revision as of 21:17, 31 January 2020
How many 
-digit positive integers (that is, integers between 
and 
, inclusive) having only even digits are divisible by 
Solution
First, we need to note that the digits have to be even, but since only one even digit for the units digit (
) we get 
.
-middletonkids
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 | ||
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