Difference between revisions of "User:Vqbc/2021 AMC 12A Problems/Problem 3"
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<math>\textbf{(A)} ~10{,}272\qquad\textbf{(B)} ~11{,}700\qquad\textbf{(C)} ~13{,}362\qquad\textbf{(D)} ~14{,}238\qquad\textbf{(E)} ~15{,}426</math> | <math>\textbf{(A)} ~10{,}272\qquad\textbf{(B)} ~11{,}700\qquad\textbf{(C)} ~13{,}362\qquad\textbf{(D)} ~14{,}238\qquad\textbf{(E)} ~15{,}426</math> | ||
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==Solution 1== | ==Solution 1== | ||
Latest revision as of 19:59, 7 November 2021
- The following problem is from both the 2021 AMC 10A #3 and 2021 AMC 12A #3, so both problems redirect to this page.
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?
Contents
- 1 Solution 1
- 2 Solution 2 (Equation)
- 3 Solution 3 (Vertical Addition and Logic)
- 4 Video Solutions
- 4.1 Video Solution (Simple)
- 4.2 Video Solution by North America Math Contest Go Go Go
- 4.3 Video Solution by Aaron He
- 4.4 Video Solution by Punxsutawney Phil
- 4.5 Video Solution by Hawk Math
- 4.6 Video Solution (Using Algebra and Meta-solving)
- 4.7 Video Solution by WhyMath
- 4.8 Video Solution by TheBeautyofMath
- 4.9 Video Solution by IceMatrix
- 4.10 Video Solution (Problems 1-3)
- 4.11 Video Solution by The Learning Royal
- 5 See also
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 (Equation)
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 
.
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
![\[\begin{array}{cccccc} & A & B & C & D & 0 \\ +\quad & & A & B & C & D \\ \hline & & & & & \\ [-2.5ex] & 1 & 7 & 4 & 0 & 2 \\ \end{array}\]](http://latex.artofproblemsolving.com/c/4/4/c44e347909afefc451d896500c564ad59cb01771.png)
Working from right to left, we get ![\[D=2\implies C=8 \implies B=5 \implies A=1.\]](http://latex.artofproblemsolving.com/b/6/1/b61dd6a92108f96ce8b9940a52e71cc0d5c86228.png)
The larger number is 
and the smaller number is 
Their difference is 
~MRENTHUSIASM
Video Solutions
Video Solution (Simple)
~ Education, the study Of Everything
Video Solution by North America Math Contest Go Go Go
https://www.youtube.com/watch?v=hMqA8i8a2SQ&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=3
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
Video Solution by WhyMath
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/50CThrk3RcM?t=107 (for AMC 10A)
Video Solution by IceMatrix
https://youtu.be/rEWS75W0Q54?t=198 (for AMC 12A)
~IceMatrix
Video Solution (Problems 1-3)
~MathWithPi
Video Solution 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. 
