AMC 8 1988 Problem 25

AMC8 1988 年第 25 题

数列与规律★★★★☆
A palindrome is a whole number that reads the same forwards and backwards. If one neglects the colon, certain times displayed on a digital watch are palindromes. Three examples are: 1:011{:}01, 4:444{:}44, and 12:2112{:}21. How many times during a 12-hour period will be palindromes?
回文数是指正读和倒读都相同的整数。如果忽略冒号,数字手表上显示的某些时间是回文数。三个例子是:1:011{:}01、4:444{:}44 和 12:2112{:}21。在 12 小时内,共有多少个时间是回文数?
  1. A.57
  2. B.60
  3. C.63
  4. D.90
  5. E.93

正确答案 · Correct Answer

A

解析 · Solution

解法1

常规解法 · 按小时位数分类 一位数小时 1–9,时刻 h:abh{:}ab 读作 hab‾\overline{hab},回文须 h=b;分钟十位 a∈{0,…,5}a\in\{0,\dots,5\} 自由,每个小时 6 个,共 9×6=54。两位数小时须 hh:mmhh{:}mm 整体回文,只有 10:0110{:}01、11:1111{:}11、12:2112{:}21 成立,3 个。合计 54+3=57,选 A。

Solution 1

Elementary · split by the hour's digits Split by the number of digits in the hour. For one-digit hours 1–9 the time h:abh{:}ab reads as hab‾\overline{hab}, a palindrome iff h=b; with the minutes' tens digit a∈{0,…,5}a\in\{0,\dots,5\} free, that is 6 per hour, so 9×6=54. For two-digit hours the whole hh:mmhh{:}mm must be a palindrome: only 10:0110{:}01, 11:1111{:}11, 12:2112{:}21 work, giving 3. The total is 54+3=57, choice (A).

继续练习

真题版权归属 MAA(AMC)。本站仅供学习用途。版权方如有异议,请联系本站予以下架。

Problems © MAA (AMC). This site is for study purposes only. Rightsholders may contact us for removal.