Difference between revisions of "2020 AMC 10A Problems/Problem 6"
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== Solution 2== | == Solution 2== | ||
The ones digit, for all the numbers that have to divisible be 5, must be a <math>0</math> or a <math>5</math>. Since the problem states that we can only use even digits, the last digit must be <math>0</math>. From there, there are no other restrictions since the divisibility rule for 5 states that the last digit must be a <math>0</math> or a <math>5</math>. So there are <math>4</math> even digit options for the first number then <math>5</math> for the middle 2. So when we have to do <math>4 \cdot 5 \cdot 5 \cdot 1 = \boxed{\textbf{(B) } 100}</math>. ~bobthefam | The ones digit, for all the numbers that have to divisible be 5, must be a <math>0</math> or a <math>5</math>. Since the problem states that we can only use even digits, the last digit must be <math>0</math>. From there, there are no other restrictions since the divisibility rule for 5 states that the last digit must be a <math>0</math> or a <math>5</math>. So there are <math>4</math> even digit options for the first number then <math>5</math> for the middle 2. So when we have to do <math>4 \cdot 5 \cdot 5 \cdot 1 = \boxed{\textbf{(B) } 100}</math>. ~bobthefam | ||
+ | == Solution 3== | ||
+ | This is basically 4*5*5 which is 100 | ||
==Video Solution== | ==Video Solution== |
Revision as of 23:39, 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
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.