Difference between revisions of "2024 AMC 8 Problems"
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Let <math>\text{x=2024}</math>. Compute the last three digits of <math>((x^3-(x-8)^3)^4-(x-69)^2)^5</math>? | Let <math>\text{x=2024}</math>. Compute the last three digits of <math>((x^3-(x-8)^3)^4-(x-69)^2)^5</math>? | ||
| − | + | It’s just a problem that just looks scary. | |
==Problem 18== | ==Problem 18== | ||
Revision as of 22:03, 20 January 2024
| 2024 AMC 8 (Answer Key) Printable versions: • AoPS Resources • PDF | ||
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Instructions
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| 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 | ||
==Problem 1== You can do it!
==Problem 1== You can do it!
Contents
- 1 Problem 3
- 2 Problem 4
- 3 Problem 5
- 4 Problem 6
- 5 Problem 8
- 6 Problem 9
- 7 Problem 10
- 8 Problem 11
- 9 Problem 12
- 10 Problem 13
- 11 Problem 14
- 12 Problem 15
- 13 Problem 16
- 14 Problem 17
- 15 Problem 18
- 16 Problem 19
- 17 Problem 20
- 18 Problem 21
- 19 Problem 22
- 20 Problem 23
- 21 Problem 24
- 22 Problem 25
- 23 See Also
Problem 3
You will not be able to cheat on this test, study for this!!
Problem 4
NOOOOO CHEATER CHEATER ONION EATER
Problem 5
bro what happened to problems 1 and 2
Problem 6
4 random points are chosen on a sphere. What is the probability that the tetrahedron with vertices of the 4 points contains the center of the sphere?
(A) 1/2 (B) 1/4 (C) 3/8 (D) 1/8 (E) 3/10 (Source: Putnam) lmao
Problem 8
Alex has 1 apple. Bob has 2 apples. How many apples did their dad eat?
(A) -1 (B) 0 (C) 1 (D) 2 (E) 3
Problem 9
Compute 
.
(A) 1 (C) 5 (B) 2 (D) 6 (E)3
Problem 10
What is the sum of the roots of 
?
A)0 B)-1 C)1 D)-2 E)2
Problem 11
The equation (2^(333x-2))+(2^(111x+2))=(2^(222x+1))+1 has three real roots. Find their sum. (Source: AIME)
You thought we could let you cheat?
Problem 12
NO CHEATERS!
Problem 13
Problem 14
Problem 15
Problem 16
Problem 17
Let. Compute the last three digits of
?
. Compute the last three digits of 
?
It’s just a problem that just looks scary.
Problem 18
hi guys. trying to cheat? im ashamed of you code: nsb
Problem 19
Problem 20
Problem 21
this question = 9+10, bc 9+10 = 21
Problem 22
What is the sum of the cubes of the solutions cubed of 
?
Problem 23
lol we are the defenders against the cheaters... get outta here and study
Problem 24
wait when are the questions coming tho
Problem 25
Did you think you could cheat the AMC ;)
See Also
| 2024 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by 2023 AMC 8 |
Followed by 2025 AMC 8 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
