Difference between revisions of "2024 AMC 8 Problems"
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==Problem 4== | ==Problem 4== | ||
− | Bob has <math>5</math> friends that go to school, only on Tuesday and Wednesday. The chance of | + | Bob has <math>5</math> friends that go to school, only on Tuesday and Wednesday. The chance of rain on Tuesday and Wednesday is <math>50\%</math>. Assuming Bob and his friends are all humans, what is the street number of Bob's house? |
<math>\textbf{(A)}\ 75423 \qquad \textbf{(B)}\ 5987 \qquad \textbf{(C)}\ 491 \qquad \textbf{(D)}\ 653\qquad \textbf{(E)}\ 21574</math> | <math>\textbf{(A)}\ 75423 \qquad \textbf{(B)}\ 5987 \qquad \textbf{(C)}\ 491 \qquad \textbf{(D)}\ 653\qquad \textbf{(E)}\ 21574</math> |
Revision as of 01:38, 21 January 2024
2024 AMC 8 (Answer Key) Printable versions: • AoPS Resources • PDF | ||
Instructions
| ||
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
If Bartholomew's pet cat's favourite number in the alphabet is purple, then what is the square root of the combined time in years Bartholomew's father went to get the milk and the time it takes for you to count to 1 million out loud.
Problem 4
Bob has friends that go to school, only on Tuesday and Wednesday. The chance of rain on Tuesday and Wednesday is
. Assuming Bob and his friends are all humans, what is the street number of Bob's house?
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
A finite set of positive integers has the property that, for each
, and each positive integer divisor
of
, there exists a unique element
satisfying
(the elements
and
could be equal).
Given this information, find all possible values for the number of elements of . (source: 2021 USAMO)
now that you read this problem you have to do it without looking at the solution or else... let's just say bad things will happen
Problem 14
Problem 15
Let be an interior point of the acute triangle
with
so that
. The point
on the segment
satisfies
, the point
on the segment
satisfies
, and the point
on the line
satisfies
. Let
and
be the circumcentres of the triangles
and
respectively. Prove that the lines
,
, and
are concurrent. (source: 2021 IMO)
now go do this problem as a punishment for trying to cheat
Problem 16
Problem 17
Let. Compute the last three digits of
?
NO CALCULATORS ARE ALLOWED.
Problem 18
hi guys. trying to cheat? im ashamed of you code: nsb
Problem 19
Write your AoPS name here if you took the AMC 8.
Probablity
Problem 20
Find the sum of the square root of -2 and the last digit of pi.
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 I think it's 1/25 for official answers since all tests end at 1/24
Problem 25
Did you think you could cheat the AMC ;)
and why did you scroll all the way here lol
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 |