2002 AMC amc12b 真题 答案 详解

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英文真题

2002 AMC 12B Problems
Problem 1
The arithmetic mean of the nine numbers in the set {9, 99, 999, 9999, ..., 999999999} is a 9-digit number M, all of whose digits are distinct. The number M does not contain the digit
(A) 0 (B) 2 (C) 4 (D) 6 (E) 8

Problem 2
What is the value of
(3x - 2)(4x + 1) - (3x - 2)4x + 1 when x = 4?
(A) 0 (B) 1 (C) 10 (D) 11 (E) 12

Problem 3
For how many positive integers n is n^2 - 3n + 2 a prime number?
(A) none (B) one (C) two (D) more than two, but finitely many (E) infinitely many

Problem 4
Let n be a positive integer such that 1/2 + 1/3 + 1/7 + 1/n is an integer. Which of the following statements is not true:
(A) 2 divides n (B) 3 divides n (C) 6 divides n (D) 7 divides n (E) n > 84

Problem 5
Let v, w, x, y, z be the degree measures of the five angles of a pentagon. Suppose that v < w < x < y < z and v, w, x, y, z form an arithmetic sequence. Find the value of x.
(A) 72 (B) 84 (C) 90 (D) 108 (E) 120

Problem 6
Suppose that a and b are nonzero real numbers, and that the equation x^2 + ax + b = 0 has solutions a and b. Then the pair (a, b) is
(A) (-2, 1) (B) (-1, 2) (C) (1, -2) (D) (2, -1) (E) (4, 4)

Problem 7
The product of three consecutive positive integers is 8 times their sum. What is the sum of their squares?
(A) 50 (B) 77 (C) 110 (D) 149 (E) 194

Problem 8
Suppose July of year N has five Mondays. Which of the following must occur five times in August of year N? (Note: Both months have 31 days.)
(A) Monday (B) Tuesday (C) Wednesday (D) Thursday (E) Friday

Problem 9
If a, b, c, d are positive real numbers such that a, b, c, d form an increasing arithmetic sequence and a, b, d form a geometric sequence, then a/d is

真题文字详解(英文)

Problem 1 The following problem is from both the 2002 AMC 12B #1 and 2002 AMC 10B #3, so both problems redirect to this page. The arithmetic mean of the nine numbers in the set {9,99,999,9999,...,999999999} is a 9-digit number M, all of whose digits are distinct. The number M doesn't contain the digit (A) 0 (B) 2 (C) 4 (D) 6 (E) 8 Solution 1 We wish to find 9+99+...+999999999 9(1+11+111+...+111111111) = 123456789 This doesn't have the digit 0, so the answer is (A) 0 Solution 2 Notice that the final number is guaranteed to have the digits {1,3,5,7,9} and that each of these digits can be paired with an even number adding up to 9. (A) 0 can be taken out, with the other digits fulfilling divisibility by 9. Solution 3 The arithmetic mean is (10^1-1)+(10^2-1)+...+(10^9-1) 9 1111111101 = 123456789 . So select A ~hastapasta Problem 2 The following problem is from both the 2002 AMC 12B #2 and 2002 AMC 10B #4, so both problems redirect to this page. What is the value of (3x - 2)(4x + 1) - (3x - 2)4x + 1 when x = 4? (A) 0 (B) 1 (C) 10 (D) 11 (E) 12 Solution 1 By the distributive property, (3x-2)[(4x+1)-4x]+1=3x-2+1=3x-1=3(4)-1=(D) 11 Solution 2 Inputting 4 into x in the original equation, [3(4)-2][4(4)+1]-[3(4)-2][4(4)]+1=170-160+1=(D) 11 See also Problem 3 The following problem is from both the 2002 AMC 12B #3 and 2002 AMC 10B #6, so both problems redirect to this page.

真题文字详解(中英双语)

Problem 1 The following problem is from both the 2002 AMC 12B #1 and 2002 AMC 10B #3, so both problems redirect to this page. 以下问题既来自2002年AMC 12B第1题，也来自2002年AMC 10B第3题，因此这两个问题都指向这一页。

The arithmetic mean of the nine numbers in the set {9,99,999,9999,...,999999999} is a 9-digit number M, all of whose digits are distinct. The number M doesn't contain the digit 集合{9,99,999,9999,...,999999999}中的九个数的算术平均数是一个9位数M，其所有数字都不相同。数字M不包含这个数字。

(A) 0 (B) 2 (C) 4 (D) 6 (E) 8

Solution 1 We wish to find 9+99+⋯+999999999 9(1+11+111+⋯+111111111) = 123456789.This doesn't have the digit 0, so the answer is (A) 0 我们希望找到 9+99+⋯+999999999 9(1+11+111+⋯+111111111) = 123456789 。这个数字不包含0，所以答案是 (A) 0

Solution 2 Notice that the final number is guaranteed to have the digits {1,3,5,7,9} and that each of these digits can be paired with an even number adding up to 9. (A) 0 can be taken out, with the other digits fulfilling divisibility by 9. 注意，最终数字保证包含数字{1,3,5,7,9}，并且这些数字中的每一个都可以与一个偶数配对，使它们的和为9。(A) 0可以被移除，其他数字满足被9整除的条件。

Solution 3 The arithmetic mean is (10^1-1)+(10^2-1)+...+(10^9-1)/9=1111111101/9=123456789.So select A 算术平均数是(10^1-1)+(10^2-1)+...+(10^9-1)/9=1111111101/9=123456789。所以选择A。

Problem 2 The following problem is from both the 2002 AMC 12B #2 and 2002 AMC 10B #4, so both problems redirect to this page. 这个问题是来自2002年AMC 12B第2题和2002年AMC 10B第4题，所以这两个问题都重定向到这个页面。

What is the value of (3x - 2)(4x + 1) - (3x - 2)4x + 1 when x = 4? 当x=4时，(3x-2)(4x+1)-(3x-2)4x+1的值是多少？

(A) 0 (B) 1 (C) 10 (D) 11 (E) 12