Difference between revisions of "2021 AMC 12A Problems/Problem 3"
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~CoolJupiter 2021 | ~CoolJupiter 2021 | ||
+ | ==Video Solution by Aaron He== | ||
+ | https://www.youtube.com/watch?v=xTGDKBthWsw&t=1m28s | ||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== | ||
https://youtube.com/watch?v=MUHja8TpKGw&t=143s | https://youtube.com/watch?v=MUHja8TpKGw&t=143s |
Revision as of 22:19, 12 February 2021
- The following problem is from both the 2021 AMC 10A #3 and 2021 AMC 12A #3, so both problems redirect to this page.
Contents
[hide]Problem
The sum of two natural numbers is . One of the two numbers is divisible by . If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
Solution 1
The units digit of a multiple of will always be . We add a whenever we multiply by . So, removing the units digit is equal to dividing by .
Let the smaller number (the one we get after removing the units digit) be . This means the bigger number would be .
We know the sum is so . So . The difference is . So, the answer is .
--abhinavg0627
Solution 2(Lazy Speed)
Since the ones place of a multiple of is , this implies the other integer has to end with a since both integers sum up to a number that ends with a . Thus, the ones place of the difference has to be , and the only answer choice that ends with an is
~CoolJupiter 2021
Video Solution by Aaron He
https://www.youtube.com/watch?v=xTGDKBthWsw&t=1m28s
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=MUHja8TpKGw&t=143s
Video Solution by Hawk Math
https://www.youtube.com/watch?v=P5al76DxyHY
Video Solution (Using Algebra and Meta-solving)
-pi_is_3.14
See also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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 |
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 2 |
Followed by Problem 4 |
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.