Difference between revisions of "2024 AMC 10A Problems/Problem 1"
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Evaluating the given expression <math>\pmod{10}</math> yields <math>1-9\equiv 2 \pmod{10}</math>, so the answer is either (A) or (D). Evaluating <math>\pmod{101}</math> yields <math>0-99\equiv 2\pmod{101}</math>. Because answer (D) is <math>202=2\cdot 101</math>, that cannot be the answer, so we choose choice <math>\boxed{\textbf{(A) }2}</math>. | Evaluating the given expression <math>\pmod{10}</math> yields <math>1-9\equiv 2 \pmod{10}</math>, so the answer is either (A) or (D). Evaluating <math>\pmod{101}</math> yields <math>0-99\equiv 2\pmod{101}</math>. Because answer (D) 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 | + | == Solution 5 (Process of Elimination) == |
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 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) }2}</math>. | 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) }2}</math>. | ||
− | + | ~[[User:Mathkiddus|mathkiddus]] | |
− | ~mathkiddus | ||
== Solution 6 (Faster Distribution) == | == Solution 6 (Faster Distribution) == |
Revision as of 20:17, 9 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
[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, fast)
- 6 Solution 5 (Process of Elimination)
- 7 Solution 6 (Faster Distribution)
- 8 Solution 7 (cubes)
- 9 Video Solution by Pi Academy
- 10 Video Solution
- 11 Video Solution 1 by Power Solve
- 12 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, fast)
Evaluating the given expression yields , so the answer is either (A) or (D). Evaluating yields . Because answer (D) 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 2 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 . ~mathkiddus
Solution 6 (Faster Distribution)
Observe that and
~laythe_enjoyer211
Solution 7 (cubes)
Let .
Then, we have
Then, the answer can be rewritten as
~erics118
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.