Difference between revisions of "2020 AMC 10A Problems/Problem 6"
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How many <math>4</math>-digit positive integers (that is, integers between <math>1000</math> and <math>9999</math>, inclusive) having only even digits are divisible by <math>5?</math> | How many <math>4</math>-digit positive integers (that is, integers between <math>1000</math> and <math>9999</math>, inclusive) having only even digits are divisible by <math>5?</math> | ||
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{{AMC10 box|year=2020|ab=A|num-b=5|num-a=7}} | {{AMC10 box|year=2020|ab=A|num-b=5|num-a=7}} | ||
+ | {{AMC12 box|year=2020|ab=A|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 07:02, 1 February 2020
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
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.