Difference between revisions of "2020 AMC 10A Problems/Problem 6"
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== Solution 3== | == Solution 3== | ||
This is basically 4*5*5 which is 100 | This is basically 4*5*5 which is 100 | ||
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==Video Solution== | ==Video Solution== |
Revision as of 23:43, 1 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.
Problem
How many -digit positive integers (that is, integers between and , inclusive) having only even digits are divisible by
Solution 1
The ones digit, for all numbers divisible by 5, must be either or . However, from the restriction in the problem, it must be even, giving us exactly one choice () for this digit. For the middle two digits, we may choose any even integer from , meaning that we have total options. For the first digit, we follow similar intuition but realize that it cannot be , hence giving us 4 possibilities. Therefore, using the multiplication rule, we get . ~ciceronii swrebby
Solution 2
The ones digit, for all the numbers that have to divisible be 5, must be a or a . Since the problem states that we can only use even digits, the last digit must be . From there, there are no other restrictions since the divisibility rule for 5 states that the last digit must be a or a . So there are even digit options for the first number then for the middle 2. So when we have to do . ~bobthefam
Solution 3
This is basically 4*5*5 which is 100
~math31415926535
Video Solution
~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.