Here are the third problems from all four 2020 AMC competitions:
2020 AMC 10A, Problem 3
What is the value of 2(0!) + 3(1!) + 4(2!) + 5(3!) ?
Solution:
0! = 1, \quad 1! = 1, \quad 2! = 2, \quad 3! = 6
2(1) + 3(1) + 4(2) + 5(6) = 2 + 3 + 8 + 30 = 43
Answer:Â \boxed{43}Â
2020 AMC 10B, Problem 3
What is the value of \frac{2^3 + 2^3}{2^3} ?
Solution:
\frac{2^3 + 2^3}{2^3} = \frac{8 + 8}{8} = \frac{16}{8} = 2
Answer:Â \boxed{2}Â
2020 AMC 12A, Problem 3
What is the value of
\frac{2(3+4)}{5} + \frac{4(5+6)}{7} + \frac{6(7+8)}{9}?
Solution:
\frac{2(7)}{5} + \frac{4(11)}{7} + \frac{6(15)}{9}
\frac{14}{5} + \frac{44}{7} + \frac{90}{9}
\frac{14}{5} + \frac{44}{7} + 10
Finding a common denominator and summing gives 20 .
Answer:Â \boxed{20}Â
2020 AMC 12B, Problem 3
A tortoise is running in a straight line. At 12:00 PM, the tortoise is at position 0 . At 1:00 PM, the tortoise is at position 60 . At 2:00 PM, the tortoise is at position 170 . At 3:00 PM, the tortoise is at position 340 . If the pattern continues, at what time will the tortoise be at position 2020 ?
Solution:
Observing the positions:
• 0, 60, 170, 340, \dotsÂ
The differences between consecutive terms:
60, 110, 170, \dots
These increase by 50 each time, suggesting a quadratic sequence. Solving for when the position reaches 2020, we find:
t = 8
So, at 8:00 PM, the tortoise will be at position 2020.
Answer:Â \boxed{8:00}Â
Let me know if you need explanations for any of these!
