Difference between revisions of "2020 AMC 10A Problems/Problem 6"
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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{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>, |
==See Also== | ==See Also== |
Revision as of 21:16, 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 ,
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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All AMC 10 Problems and Solutions |
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