Difference between revisions of "2021 AMC 12A Problems/Problem 3"
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− | {{duplicate|[[2021 AMC 10A Problems | + | {{duplicate|[[2021 AMC 10A Problems/Problem 3|2021 AMC 10A #3]] and [[2021 AMC 12A Problems/Problem 3|2021 AMC 12A #3]]}} |
==Problem== | ==Problem== | ||
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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 1== | + | ==Solution 1 (Algebra)== |
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>. | ||
Line 12: | Line 12: | ||
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) | + | 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) = \boxed{\textbf{(D)} ~14{,}238}</math>. |
+ | ~abhinavg0627 | ||
− | + | ==Solution 2 (Arithmetic)== | |
− | = | + | Since the unit's place of a multiple of <math>10</math> is <math>0</math>, the other integer must end with a <math>2</math>, for both integers sum up to a number ending in a <math>2</math>. Thus, the unit's place of the difference must be <math>10-2=8</math>, and the only answer choice that ends with an <math>8</math> is <math>\boxed{\textbf{(D)} ~14{,}238}</math>. |
− | + | Another quick solution is to realize that the sum 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. | |
~CoolJupiter 2021 | ~CoolJupiter 2021 | ||
− | + | ~Extremelysupercooldude (Minor grammar edits) | |
==Solution 3 (Vertical Addition and Logic)== | ==Solution 3 (Vertical Addition and Logic)== | ||
− | Let the larger number be <math>\ | + | 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} | <cmath>\begin{array}{cccccc} | ||
& A & B & C & D & 0 \\ | & A & B & C & D & 0 \\ | ||
− | + & & A & B & C & D \\ | + | +\quad & & A & B & C & D \\ |
\hline | \hline | ||
+ | & & & & & \\ [-2.5ex] | ||
& 1 & 7 & 4 & 0 & 2 \\ | & 1 & 7 & 4 & 0 & 2 \\ | ||
\end{array}</cmath> | \end{array}</cmath> | ||
− | Working from right to left, we | + | 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> | + | 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 | ~MRENTHUSIASM | ||
− | ==Video Solution by North America Math Contest Go Go Go== | + | ==Solution 4 (Logic)== |
+ | We know that the larger number has a units digit of <math>0</math> since it is divisible by 10. If <math>D</math> is the ten's digit of the larger number, then <math>D</math> is the units digit of the smaller number. Since the sum of the natural numbers has a unit's digit of <math>2</math>, <math>D=2</math>. | ||
+ | |||
+ | The units digit of the larger number is <math>0</math> and the units digit of the smaller number is <math>2</math>, so the positive difference between the numbers is 8. There is only one answer choice that has this units digit, and that is <math>\boxed{\textbf{(D)} ~14{,}238}.</math> | ||
+ | |||
+ | (Similar to MRENTHUSIASM's solution) | ||
+ | |||
+ | ~AARUSHGORADIA18 | ||
+ | |||
+ | ==Video Solutions== | ||
+ | |||
+ | ==Video Solution 1 (easily understandable)== | ||
+ | https://www.youtube.com/watch?v=jPzcN0KeJ88&t=132s | ||
+ | |||
+ | ===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 | https://www.youtube.com/watch?v=hMqA8i8a2SQ&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=3 | ||
− | ==Video Solution by Aaron He== | + | ===Video Solution by Aaron He=== |
https://www.youtube.com/watch?v=xTGDKBthWsw&t=1m28s | 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 | ||
− | ==Video Solution by Hawk Math== | + | ===Video Solution by Hawk Math=== |
https://www.youtube.com/watch?v=P5al76DxyHY | https://www.youtube.com/watch?v=P5al76DxyHY | ||
− | ==Video Solution (Using Algebra and Meta-solving)== | + | ===Video Solution by OmegaLearn (Using Algebra and Meta-solving)=== |
https://youtu.be/d2musztzDjw | https://youtu.be/d2musztzDjw | ||
-pi_is_3.14 | -pi_is_3.14 | ||
− | ==Video Solution by WhyMath== | + | ===Video Solution by WhyMath=== |
https://youtu.be/VpYmQEKcBpA | https://youtu.be/VpYmQEKcBpA | ||
~savannahsolver | ~savannahsolver | ||
− | ==Video Solution by TheBeautyofMath== | + | ===Video Solution by TheBeautyofMath=== |
https://youtu.be/50CThrk3RcM?t=107 (for AMC 10A) | https://youtu.be/50CThrk3RcM?t=107 (for AMC 10A) | ||
+ | ===Video Solution by IceMatrix=== | ||
https://youtu.be/rEWS75W0Q54?t=198 (for AMC 12A) | https://youtu.be/rEWS75W0Q54?t=198 (for AMC 12A) | ||
~IceMatrix | ~IceMatrix | ||
− | ==Video Solution (Problems 1-3)== | + | ===Video Solution (Problems 1-3)=== |
https://youtu.be/CupJpUzKPB0 | https://youtu.be/CupJpUzKPB0 | ||
~MathWithPi | ~MathWithPi | ||
+ | |||
+ | ===Video Solution by The Learning Royal=== | ||
+ | https://youtu.be/slVBYmcDMOI | ||
==See also== | ==See also== |
Latest revision as of 21:35, 26 December 2023
- 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 (Algebra)
- 3 Solution 2 (Arithmetic)
- 4 Solution 3 (Vertical Addition and Logic)
- 5 Solution 4 (Logic)
- 6 Video Solutions
- 7 Video Solution 1 (easily understandable)
- 7.1 Video Solution (Simple)
- 7.2 Video Solution by North America Math Contest Go Go Go
- 7.3 Video Solution by Aaron He
- 7.4 Video Solution by Punxsutawney Phil
- 7.5 Video Solution by Hawk Math
- 7.6 Video Solution by OmegaLearn (Using Algebra and Meta-solving)
- 7.7 Video Solution by WhyMath
- 7.8 Video Solution by TheBeautyofMath
- 7.9 Video Solution by IceMatrix
- 7.10 Video Solution (Problems 1-3)
- 7.11 Video Solution by The Learning Royal
- 8 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 (Algebra)
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 (Arithmetic)
Since the unit's place of a multiple of is
, the other integer must end with a
, for both integers sum up to a number ending in a
. Thus, the unit's place of the difference must be
, and the only answer choice that ends with an
is
.
Another quick solution is to realize that the sum represents a number added to
. The difference is
, which is
of the given sum.
~CoolJupiter 2021
~Extremelysupercooldude (Minor grammar edits)
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
Solution 4 (Logic)
We know that the larger number has a units digit of since it is divisible by 10. If
is the ten's digit of the larger number, then
is the units digit of the smaller number. Since the sum of the natural numbers has a unit's digit of
,
.
The units digit of the larger number is and the units digit of the smaller number is
, so the positive difference between the numbers is 8. There is only one answer choice that has this units digit, and that is
(Similar to MRENTHUSIASM's solution)
~AARUSHGORADIA18
Video Solutions
Video Solution 1 (easily understandable)
https://www.youtube.com/watch?v=jPzcN0KeJ88&t=132s
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 by OmegaLearn (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.