Difference between revisions of "2020 AMC 10A Problems/Problem 6"
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== Solution == | == Solution == | ||
The ones digit, for all numbers divisible by 5, must be either <math>0</math> or <math>5</math>. However, from the restriction in the problem, it must be even, giving us exactly one choice (<math>0</math>) for this digit. For the middle two digits, we may choose any even integer from <math>[0, 8]</math>, meaning that we have <math>5</math> total options. For the first digit, we follow similar intuition but realize that it cannot be <math>0</math>, hence giving us 4 possibilities. Therefore, using the multiplication rule, we get <math>4\times 5 \times 5 \times 1 = \boxed{\textbf{(B) } 100}</math>. ~ciceronii | The ones digit, for all numbers divisible by 5, must be either <math>0</math> or <math>5</math>. However, from the restriction in the problem, it must be even, giving us exactly one choice (<math>0</math>) for this digit. For the middle two digits, we may choose any even integer from <math>[0, 8]</math>, meaning that we have <math>5</math> total options. For the first digit, we follow similar intuition but realize that it cannot be <math>0</math>, hence giving us 4 possibilities. Therefore, using the multiplication rule, we get <math>4\times 5 \times 5 \times 1 = \boxed{\textbf{(B) } 100}</math>. ~ciceronii | ||
+ | swrebby | ||
==Video Solution== | ==Video Solution== |
Revision as of 16:55, 18 March 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
[hide]Problem
How many -digit positive integers (that is, integers between
and
, inclusive) having only even digits are divisible by
Solution
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
Video Solution
~IceMatrix
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.