2021 AMC 12A Problems/Problem 3
- 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
Another quick solution is to realize that the sum is represents a number added to . The difference is , which is of the given sum.
Solution 3 (Vertical Addition and Logic)
Let the larger number be It follows that the smaller number is Adding vertically, we have Working from right to left, we get The larger number is and the smaller number is Their difference is
~MRENTHUSIASM
Video Solutions
(Simple)
~ Education, the study Of Everything
by North America Math Contest Go Go Go
https://www.youtube.com/watch?v=hMqA8i8a2SQ&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=3
by Aaron He
https://www.youtube.com/watch?v=xTGDKBthWsw&t=1m28s
by Punxsutawney Phil
https://youtube.com/watch?v=MUHja8TpKGw&t=143s
by Hawk Math
https://www.youtube.com/watch?v=P5al76DxyHY
(Using Algebra and Meta-solving)
-pi_is_3.14
by WhyMath
~savannahsolver
by TheBeautyofMath
https://youtu.be/50CThrk3RcM?t=107 (for AMC 10A)
by IceMatrix
https://youtu.be/rEWS75W0Q54?t=198 (for AMC 12A)
~IceMatrix
(Problems 1-3)
~MathWithPi
by The Learning Royal
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.