AMC12 2010 B

AMC12 2010 B · Q11

AMC12 2010 B · Q11. It mainly tests Probability (basic), Divisibility & factors.

A palindrome between $1000$ and $10,000$ is chosen at random. What is the probability that it is divisible by $7$?

在 $1000$ 到 $10,000$ 之间随机选择一个回文数。它能被 $7$ 整除的概率是多少？

(A) \frac{1}{10} \frac{1}{10}

(B) \frac{1}{9} \frac{1}{9}

(D) \frac{1}{6} \frac{1}{6}

(E) \frac{1}{5} \frac{1}{5}

Answer

Correct choice: (E)

正确答案：（E）

Solution

View the palindrome as some number with form (decimal representation): $a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$. But because the number is a palindrome, $a_3 = a_0, a_2 = a_1$. Recombining this yields $1001a_3 + 110a_2$. 1001 is divisible by 7, which means that as long as $a_2 = 0$, the palindrome will be divisible by 7. This yields 9 palindromes out of 90 ($9 \cdot 10$) possibilities for palindromes. However, if $a_2 = 7$, then this gives another case in which the palindrome is divisible by 7. This adds another 9 palindromes to the list, bringing our total to $18/90 = \boxed {\frac{1}{5} } = \boxed {E}$

将回文数看作（十进制表示）形如 $a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$ 的数。但因为该数是回文数，$a_3 = a_0, a_2 = a_1$。重新组合可得 $1001a_3 + 110a_2$。$1001$ 能被 $7$ 整除，这意味着只要 $a_2 = 0$，该回文数就能被 $7$ 整除。这在 $90$（$9 \cdot 10$）个可能的回文数中给出 $9$ 个回文数。然而，如果 $a_2 = 7$，也会出现该回文数能被 $7$ 整除的情况。这又增加了 $9$ 个回文数，使总数为 $18/90 = \boxed {\frac{1}{5} } = \boxed {E}$

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