Difference between revisions of "2020 AMC 10A Problems/Problem 6"
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== Solution 2 == | == Solution 2 == | ||
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 (0) 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 (0) 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 | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/JEjib74EmiY | ||
| + | |||
| + | ~IceMatrix | ||
==See Also== | ==See Also== | ||
Revision as of 22:11, 31 January 2020
How many 
-digit positive integers (that is, integers between 
and 
, inclusive) having only even digits are divisible by 
Contents
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 
.
-middletonkids
Solution 2
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 (0) for this digit. For the middle two digits, we may choose any even integer from ![$[0, 8]$](http://latex.artofproblemsolving.com/5/1/5/51503c51fecd7d489a3c90ac1c108785aa607446.png)
, 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
