Difference between revisions of "2021 AMC 12A Problems/Problem 3"
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+ | {{duplicate|[[2021 AMC 10A Problems#Problem 3|2021 AMC 10A #3]] and [[2021 AMC 12A Problems#Problem 3|2021 AMC 12A #3]]}} | ||
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==Problem== | ==Problem== | ||
The sum of two natural numbers is <math>17{,}402</math>. One of the two numbers is divisible by <math>10</math>. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers? | The sum of two natural numbers is <math>17{,}402</math>. One of the two numbers is divisible by <math>10</math>. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers? | ||
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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> | ||
− | ==Solution== | + | ==Solution 1== |
The units digit of a multiple of <math>10</math> will always be <math>0</math>. We add a <math>0</math> whenever we multiply by <math>10</math>. So, removing the units digit is equal to dividing by <math>10</math>. | The units digit of a multiple of <math>10</math> will always be <math>0</math>. We add a <math>0</math> whenever we multiply by <math>10</math>. So, removing the units digit is equal to dividing by <math>10</math>. | ||
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Let the smaller number (the one we get after removing the units digit) be <math>a</math>. This means the bigger number would be <math>10a</math>. | Let the smaller number (the one we get after removing the units digit) be <math>a</math>. This means the bigger number would be <math>10a</math>. | ||
− | We know the sum is <math>10a+a = 11a</math> so <math>11a=17402</math>. So <math>a=1582</math>. The difference is <math>10a-a = 9a</math>. So, the answer is <math>9(1582) = 14238</math>. | + | We know the sum is <math>10a+a = 11a</math> so <math>11a=17402</math>. So <math>a=1582</math>. The difference is <math>10a-a = 9a</math>. So, the answer is <math>9(1582) = 14238 = \boxed{\textbf{(D)}}</math>. |
+ | |||
+ | --abhinavg0627 | ||
+ | |||
+ | ==Solution 2 (Lazy Speed)== | ||
+ | |||
+ | Since the ones place of a multiple of <math>10</math> is <math>0</math>, this implies the other integer has to end with a <math>2</math> since both integers sum up to a number that ends with a <math>2</math>. Thus, the ones place of the difference has to be <math>10-2=8</math>, and the only answer choice that ends with an <math>8</math> is <math>\boxed{\textbf{(D)}~14238}</math> | ||
+ | |||
+ | ~CoolJupiter 2021 | ||
+ | |||
+ | Another quick solution is to realize that the sum is represents a number <math>n</math> added to <math>10n</math>. The difference is <math>9n</math>, which is <math>\frac{9}{11}</math> of the given sum. | ||
+ | |||
+ | ==Solution 3 (Vertical Addition and Logic)== | ||
+ | Let the larger number be <math>\underline{AB,CD0}.</math> It follows that the smaller number is <math>\underline{A,BCD}.</math> Adding vertically, we have | ||
+ | <cmath>\begin{array}{cccccc} | ||
+ | & A & B & C & D & 0 \\ | ||
+ | + & & A & B & C & D \\ | ||
+ | \hline | ||
+ | & & & & & \\ [-2.5ex] | ||
+ | & 1 & 7 & 4 & 0 & 2 \\ | ||
+ | \end{array}</cmath> | ||
+ | Working from right to left, we get <cmath>D=2\implies C=8 \implies B=5 \implies A=1.</cmath> | ||
+ | The larger number is <math>15,820</math> and the smaller number is <math>1,582.</math> Their difference is <math>15,820-1,582=\boxed{\textbf{(D)} ~14{,}238}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==Video Solution (Simple)== | ||
+ | https://youtu.be/SEp9flDYm2c | ||
+ | |||
+ | ~ 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)== | ||
+ | https://youtu.be/d2musztzDjw | ||
+ | |||
+ | -pi_is_3.14 | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/VpYmQEKcBpA | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/50CThrk3RcM?t=107 (for AMC 10A) | ||
+ | |||
+ | https://youtu.be/rEWS75W0Q54?t=198 (for AMC 12A) | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==Video Solution (Problems 1-3)== | ||
+ | https://youtu.be/CupJpUzKPB0 | ||
+ | |||
+ | ~MathWithPi | ||
+ | |||
+ | ==Video Solution by The Learning Royal== | ||
+ | https://youtu.be/slVBYmcDMOI | ||
==See also== | ==See also== | ||
+ | {{AMC10 box|year=2021|ab=A|num-b=2|num-a=4}} | ||
{{AMC12 box|year=2021|ab=A|num-b=2|num-a=4}} | {{AMC12 box|year=2021|ab=A|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:30, 3 July 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
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Lazy Speed)
- 4 Solution 3 (Vertical Addition and Logic)
- 5 Video Solution (Simple)
- 6 Video Solution by North America Math Contest Go Go Go
- 7 Video Solution by Aaron He
- 8 Video Solution by Punxsutawney Phil
- 9 Video Solution by Hawk Math
- 10 Video Solution (Using Algebra and Meta-solving)
- 11 Video Solution by WhyMath
- 12 Video Solution by TheBeautyofMath
- 13 Video Solution (Problems 1-3)
- 14 Video Solution by The Learning Royal
- 15 See also
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 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)
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.