2021 AMC FALL amc12b 真题 答案 详解

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中英双语真题

2021Fall AMC12B
Problem 1
What is the value of 1234 + 2341 + 3412 + 4123?
表达式 1234 + 2341 + 3412 + 4123 的值是多少？
(A) 10,000 (B) 10,010 (C) 10,110 (D) 11,000 (E) 11,110

Problem 2
What is the area of the shaded figure shown below?
下图中阴影部分的面积是多少？
(A) 4 (B) 6 (C) 8 (D) 10 (E) 12

英文真题

2021 Fall AMC 12B Problems Problem 1 What is the value of $1234 + 2341 + 3412 + 4123$?
(A) 10,000 (B) 10,010 (C) 10,110 (D) 11,000 (E) 11,110

Problem 2 What is the area of the shaded figure shown below?
(A) 4 (B) 6 (C) 8 (D) 10 (E) 12

Problem 3 At noon on a certain day, Minneapolis is $N$ degrees warmer than St. Louis. At 4:00 the temperature in Minneapolis has fallen by 5 degrees while the temperature in St. Louis has risen by 3 degrees, at which time the temperatures in the two cities differ by 2 degrees. What is the product of all possible values of $N$?
(A) 10 (B) 30 (C) 60 (D) 100 (E) 120

Problem 4 Let $n = 8^{2022}$. Which of the following is equal to $\frac{n}{4}$?
(A) $4^{1010}$ (B) $2^{2022}$ (C) $8^{2018}$ (D) $4^{3031}$ (E) $4^{3032}$

Problem 5 Call a fraction $\frac{a}{b}$, not necessarily in the simplest form, special if $a$ and $b$ are positive integers whose sum is 15. How many distinct integers can be written as the sum of two, not necessarily different, special fractions?
(A) 9 (B) 10 (C) 11 (D) 12 (E) 13

Problem 6 The greatest prime number that is a divisor of 16,384 is 2 because $16,384 = 2^{14}$. What is the sum of the digits of the greatest prime number that is a divisor of 16,383?
(A) 3 (B) 7 (C) 10 (D) 16 (E) 22

真题文字详解(英文)

Problem1 What is the value of 1234 + 2341 + 3412 + 4123 (A) 10,000 (B) 10,010 (C) 10,110 (D) 11,000 (E) 11,110 Solution 1 We see that 1, 2, 3, and 4 each appear in the ones, tens, hundreds, and thousands digit exactly once. Since 1 + 2 + 3 + 4 = 10, we find that the sum is equal to 10 · (1 + 10 + 100 + 1000) = (E) 11,110 Note that it is equally valid to manually add all four numbers together to get the answer. Solution 2 We have 1234 + 2341 + 3412 + 4123 = 1111(1 + 2 + 3 + 4) = (E) 11,110 ~Steven Chen (www.professorchenedu.com) Solution 3 We see that the units digit must be 0, since 4 + 3 + 2 + 1 is 0. But every digit from there will be a 1 since we have that each time afterwards, we must carry the 1 from the previous sum. The answer choice that satisfies these conditions is (E) 11,110 Solution 4 (Brute Force) We can simply add the numbers. 1234 + 2341 + 3412 + 4123 = 11110 => (E) Note: Although this would not take terribly long, it is not recommended to do this in a real contest. For AMC 10: https://youtu.be/lc7naDZ1Eu4 For AMC 12: https://youtu.be/yaE5aAmesc

真题文字详解(中英双语)

Problem 1 The following problem is from both the 2021 Fall AMC 10B #1 and 2021 Fall AMC 12B #1, so both problems redirect to this page. What is the value of $1234 + 2341 + 3412 + 4123$?
(A) 10,000 (B) 10,010 (C) 10,110 (D) 11,000 (E) 11,110 Solution 1 We see that 1, 2, 3, and 4 each appear in the ones, tens, hundreds, and thousands digit exactly once. Since $1+2+3+4=10$, we find that the sum is equal to $10\cdot(1+10+100+1000)=\boxed{\text{(E)} 11,110}$. Note that it is equally valid to manually add all four numbers together to get the answer. Solution 2 We have $1234+2341+3412+4123=1111(1+2+3+4)=\boxed{\text{(E)} 11,110}.$ Solution 3 We see that the units digit must be 0, since $4+3+2+1=0$. But every digit from there will be a 1 since we have that each time afterwards, we must carry the 1 from the previous sum. The answer choice that satisfies these conditions is $\boxed{\text{(E)} 11,110}$. Problem 2 The following problem is from both the 2021 Fall AMC 10B #2 and 2021 Fall AMC 12B #2, so both problems redirect to this page. What is the area of the shaded figure shown below? (A) 4 (B) 6 (C) 8 (D) 10 (E) 12 Solution 1 (Area Addition)