Difference between revisions of "2024 AMC 10A Problems/Problem 1"
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== Problem == | == Problem == | ||
− | What is the value of <math> | + | What is the value of <math>9901\cdot101-99\cdot10101?</math> |
− | <math>\textbf{(A)}~2\qquad\textbf{(B)}~ | + | <math>\textbf{(A)}~2\qquad\textbf{(B)}~20\qquad\textbf{(C)}~200\qquad\textbf{(D)}~202\qquad\textbf{(E)}~2020</math> |
− | + | == Solution 1 (Direct Computation) == | |
− | + | The likely fastest method will be direct computation. <math>9901\cdot101</math> evaluates to <math>1000001</math> and <math>99\cdot10101</math> evaluates to <math>999999</math>. The difference is <math>\boxed{\textbf{(A) }2}.</math> | |
− | + | Solution by [[User:Juwushu|juwushu]]. | |
+ | |||
+ | == Solution 2 (Distributive Property) == | ||
+ | We have | ||
+ | <cmath>\begin{align*} | ||
+ | 9901\cdot101-99\cdot10101 &= (10000-99)\cdot101-99\cdot(10000+101) \ | ||
+ | &= 10000\cdot101-99\cdot101-99\cdot10000-99\cdot101 \ | ||
+ | &= (10000\cdot101-99\cdot10000)-2\cdot(99\cdot101) \ | ||
+ | &= 2\cdot10000-2\cdot9999 \ | ||
+ | &= \boxed{\textbf{(A) }2}. | ||
+ | \end{align*}</cmath> | ||
+ | ~MRENTHUSIASM | ||
== Solution 3 (Solution 1 but Distributive) == | == Solution 3 (Solution 1 but Distributive) == | ||
+ | Note that <math>9901\cdot101=9901\cdot100+9901=990100+9901=1000001</math> and <math>99\cdot10101=100\cdot10101-10101=1010100-10101=999999</math>, therefore the answer is <math>1000001-999999=\boxed{\textbf{(A) }2}</math>. | ||
− | + | ~Tacos_are_yummy_1 | |
− | |||
== Solution 4 (Modular Arithmetic) == | == Solution 4 (Modular Arithmetic) == | ||
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We simply look at the units digit of the problem we have (or take mod <math>10</math>) | We simply look at the units digit of the problem we have (or take mod <math>10</math>) | ||
<cmath>9901\cdot101-99\cdot10101 \equiv 1\cdot1 - 9\cdot1 = 2 \mod{10}.</cmath> | <cmath>9901\cdot101-99\cdot10101 \equiv 1\cdot1 - 9\cdot1 = 2 \mod{10}.</cmath> | ||
− | Since the only answer with <math>2</math> in the units digit is <math>\textbf{(A)}</math> | + | Since the only answer with <math>2</math> in the units digit is <math>\textbf{(A)}</math>, We can then continue if you are desperate to use guess and check or a actually valid method to find the answer is <math>\boxed{\textbf{(A) }2}</math>. |
~[[User:Mathkiddus|mathkiddus]] | ~[[User:Mathkiddus|mathkiddus]] | ||
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Let <math>x=100</math>. Then, we have | Let <math>x=100</math>. Then, we have | ||
\begin{align*} | \begin{align*} | ||
− | 101\cdot 9901=(x+1)\cdot (x^2-x+1)=x^3+1 \ | + | 101\cdot 9901=(x+1)\cdot (x^2-x+1)=x^3+1, \ |
− | 99\cdot 10101=(x-1)\cdot (x^2+x+1)=x^3-1 | + | 99\cdot 10101=(x-1)\cdot (x^2+x+1)=x^3-1. |
\end{align*} | \end{align*} | ||
− | + | Then, the answer can be rewritten as <math>(x^3+1)-(x^3-1)= \boxed{\textbf{(A) }2}.</math> | |
− | Then, the answer can be rewritten as <math>(x^3+1)-(x^3-1)=\boxed{\textbf{(A) }2}</math> | ||
~erics118 | ~erics118 | ||
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~RULE101 | ~RULE101 | ||
+ | |||
+ | == Video Solution (⚡️ 1 min solve ⚡️) == | ||
+ | |||
+ | https://youtu.be/RODYXdpipdc | ||
+ | |||
+ | <i>~Education, the Study of Everything </i> | ||
== Video Solution by Pi Academy == | == Video Solution by Pi Academy == | ||
https://youtu.be/GPoTfGAf8bc?si=JYDhLVzfHUbXa3DW | https://youtu.be/GPoTfGAf8bc?si=JYDhLVzfHUbXa3DW | ||
+ | |||
+ | == Video Solution by FrankTutor == | ||
+ | https://www.youtube.com/watch?v=ez095SvW5xI | ||
== Video Solution Daily Dose of Math == | == Video Solution Daily Dose of Math == |
Revision as of 10:58, 29 December 2024
- The following problem is from both the 2024 AMC 10A #1 and 2024 AMC 12A #1, so both problems redirect to this page.
Contents
[hide]- 1 Problem
- 2 Solution 1 (Direct Computation)
- 3 Solution 2 (Distributive Property)
- 4 Solution 3 (Solution 1 but Distributive)
- 5 Solution 4 (Modular Arithmetic)
- 6 Solution 5 (Process of Elimination)
- 7 Solution 6 (Faster Distribution)
- 8 Solution 7 (Cubes)
- 9 Solution 8 (Super Fast)
- 10 Video Solution (⚡️ 1 min solve ⚡️)
- 11 Video Solution by Pi Academy
- 12 Video Solution by FrankTutor
- 13 Video Solution Daily Dose of Math
- 14 Video Solution 1 by Power Solve
- 15 Video Solution by SpreadTheMathLove
- 16 See also
Problem
What is the value of
Solution 1 (Direct Computation)
The likely fastest method will be direct computation. evaluates to
and
evaluates to
. The difference is
Solution by juwushu.
Solution 2 (Distributive Property)
We have
~MRENTHUSIASM
Solution 3 (Solution 1 but Distributive)
Note that and
, therefore the answer is
.
~Tacos_are_yummy_1
Solution 4 (Modular Arithmetic)
Evaluating the given expression yields
, so the answer is either
or
. Evaluating
yields
. Because answer
is
, that cannot be the answer, so we choose choice
.
Solution 5 (Process of Elimination)
We simply look at the units digit of the problem we have (or take mod )
Since the only answer with
in the units digit is
, We can then continue if you are desperate to use guess and check or a actually valid method to find the answer is
.
Solution 6 (Faster Distribution)
Observe that and
~laythe_enjoyer211
Solution 7 (Cubes)
Let . Then, we have
~erics118
Solution 8 (Super Fast)
It's not hard to observe and express into
, and
into
.
We then simplify the original expression into , which could then be simplified into
, which we can get the answer of
.
~RULE101
Video Solution (⚡️ 1 min solve ⚡️)
~Education, the Study of Everything
Video Solution by Pi Academy
https://youtu.be/GPoTfGAf8bc?si=JYDhLVzfHUbXa3DW
Video Solution by FrankTutor
https://www.youtube.com/watch?v=ez095SvW5xI
Video Solution Daily Dose of Math
~Thesmartgreekmathdude
Video Solution 1 by Power Solve
https://www.youtube.com/watch?v=j-37jvqzhrg
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=6SQ74nt3ynw
See also
2024 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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 |
2024 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by First Problem |
Followed by Problem 2 |
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.