AMC10 2009 A
AMC10 2009 A · Q9
AMC10 2009 A · Q9. It mainly tests Primes & prime factorization.
Positive integers $a$, $b$, and 2009, with $a < b < 2009$, form a geometric sequence with an integer ratio. What is $a$?
正整数 $a$、$b$ 和 2009 构成公比为整数的等比数列，且 $a < b < 2009$。$a$ 等于多少？
(A)
7
7
(B)
41
41
(C)
49
49
(D)
287
287
(E)
2009
2009
Answer
Correct choice: (B)
正确答案：（B）
Solution
Answer (B): Let the ratio be $r$. Then $ar^2 = 2009 = 41\cdot 7^2$. Because $r$ must be an integer greater than 1, the only possible value of $r$ is 7, and $a = 41$.
答案（B）：设公比为 $r$。则 $ar^2 = 2009 = 41\cdot 7^2$。由于 $r$ 必须是大于 1 的整数，$r$ 的唯一可能取值为 7，因此 $a = 41$。
Topics
Related Questions
Practice full AMC exams on amcdrill.
Try full-length practice and diagnostics at www.amcdrill.com.