Difference between revisions of "2024 AMC 10A Problems/Problem 1"
MRENTHUSIASM (talk | contribs) (→Solution 4 (Modular Arithmetic)) |
|||
(22 intermediate revisions by 9 users not shown) | |||
Line 3: | Line 3: | ||
== Problem == | == Problem == | ||
− | What is the value of < | + | What is the value of <math>9901\cdot101-99\cdot10101?</math> |
<math>\textbf{(A)}~2\qquad\textbf{(B)}~20\qquad\textbf{(C)}~200\qquad\textbf{(D)}~202\qquad\textbf{(E)}~2020</math> | <math>\textbf{(A)}~2\qquad\textbf{(B)}~20\qquad\textbf{(C)}~200\qquad\textbf{(D)}~202\qquad\textbf{(E)}~2020</math> | ||
Line 22: | Line 22: | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | |||
− | We simply look at the units digit of the problem we have(or take mod 10) | + | == Solution 3 (Solution 1 but Distributive) == |
+ | Note that <math>9901\cdot101=9901\cdot100+9901=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) == | ||
+ | Evaluating the given expression <math>\pmod{10}</math> yields <math>1-9\equiv 2 \pmod{10}</math>, so the answer is either <math>\textbf{(A)}</math> or <math>\textbf{(D)}</math>. Evaluating <math>\pmod{101}</math> yields <math>0-99\equiv 2\pmod{101}</math>. Because answer <math>\textbf{(D)}</math> is <math>202=2\cdot 101</math>, that cannot be the answer, so we choose choice <math>\boxed{\textbf{(A) }2}</math>. | ||
+ | |||
+ | == Solution 5 (Process of Elimination) == | ||
+ | |||
+ | 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 2 in the units digit is <math>\textbf{(A)}</math> or <math>\textbf{(D)}</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)}}</math> | + | Since the only answer with <math>2</math> in the units digit is <math>\textbf{(A)}</math> or <math>\textbf{(D)}</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>. |
− | ~mathkiddus | + | |
− | + | ~[[User:Mathkiddus|mathkiddus]] | |
− | |||
− | == Video Solution by | + | == Solution 6 (Faster Distribution) == |
+ | Observe that <math>9901=9900+1=99\cdot100+1</math> and <math>10101=10100+1=101\cdot100+1</math> | ||
+ | <cmath>\begin{align*} | ||
+ | \Rightarrow9901\cdot101-99\cdot10101 & = ((9900\cdot101)+(1\cdot101))-((99\cdot10100)+(99\cdot1)) \\ | ||
+ | &=(99\cdot100\cdot101)+101-(99\cdot100\cdot101)-99 \\ | ||
+ | &=101-99 \\ | ||
+ | &=\boxed{\textbf{(A) }2}. | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | ~laythe_enjoyer211 | ||
+ | |||
+ | ==Solution 7 (Cubes)== | ||
+ | |||
+ | Let <math>x=100</math>. Then, we have | ||
+ | \begin{align*} | ||
+ | 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 | ||
+ | \end{align*} | ||
+ | |||
+ | Then, the answer can be rewritten as <math>(x^3+1)-(x^3-1)=\boxed{\textbf{(A) }2}</math> | ||
+ | |||
+ | ~erics118 | ||
+ | |||
+ | ==Solution 8 (Super Fast)== | ||
+ | |||
+ | It's not hard to observe and express <math>9901</math> into <math>99\cdot100+1</math>, and <math>10101</math> into <math>101\cdot100+1</math>. | ||
+ | |||
+ | We then simplify the original expression into <math>(99\cdot100+1)\cdot101-99\cdot(101\cdot100+1)</math>, which could then be simplified into <math>99\cdot100\cdot101+101-99\cdot100\cdot101-99</math>, which we can get the answer of <math>101-99=\boxed{\textbf{(A) }2}</math>. | ||
+ | |||
+ | ~RULE101 | ||
+ | |||
+ | == Video Solution by Pi Academy == | ||
+ | https://youtu.be/GPoTfGAf8bc?si=JYDhLVzfHUbXa3DW | ||
+ | |||
+ | == Video Solution == | ||
https://youtu.be/Z76bafQsqTc | https://youtu.be/Z76bafQsqTc | ||
~Thesmartgreekmathdude | ~Thesmartgreekmathdude | ||
+ | |||
+ | == Video Solution 1 by Power Solve == | ||
+ | https://www.youtube.com/watch?v=j-37jvqzhrg | ||
==See also== | ==See also== |
Latest revision as of 10:13, 11 November 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
- 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 by Pi Academy
- 11 Video Solution
- 12 Video Solution 1 by Power Solve
- 13 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 or 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 \begin{align*} 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 \end{align*}
Then, the answer can be rewritten as
~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 by Pi Academy
https://youtu.be/GPoTfGAf8bc?si=JYDhLVzfHUbXa3DW
Video Solution
~Thesmartgreekmathdude
Video Solution 1 by Power Solve
https://www.youtube.com/watch?v=j-37jvqzhrg
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.