1988 AMC 8 Problem 25

Attempt Problem 25 of the 1988 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1988 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

25.

A palindrome is a whole number that reads the same forwards as backwards. If one neglects the colon, certain times displayed on a digital watch are palindromes. Three examples are 1:01,1{:}01, 4:44,4{:}44, and 12:21.12{:}21. How many times during a 1212-hour period will be palindromes?
5757
6060
6363
9090
9393
Answer: A
Concepts:palindromecasework
Small hint:
Split into single-digit hours (11 through 99) and two-digit hours (10,10, 11,11, 1212), since the number of digits differs
Big hint:
For a single-digit hour h,h, the time h ⁣: ⁣m1m2h\!:\!m_1 m_2 is a palindrome when m2=h,m_2 = h, leaving m1m_1 free among 00 through 55
Solution:
For a single-digit hour hh (from 11 to 99), the display h ⁣: ⁣m1m2h\!:\!m_1 m_2 reads the same backwards when m2=h.m_2 = h. The tens digit m1m_1 can be 0,0, 1,1, 2,2, 3,3, 4,4, or 5,5, giving 66 palindromes per hour, so 9×6=54.9 \times 6 = 54.
For the two-digit hours 10,10, 11,11, 12,12, the display h1h2 ⁣: ⁣m1m2h_1 h_2\!:\!m_1 m_2 is a palindrome when m1=h2m_1 = h_2 and m2=h1,m_2 = h_1, giving exactly 10:01,10{:}01, 11:11,11{:}11, and 12:2112{:}21 — one each, for 33 more.
The total is 54+3=57.54 + 3 = 57.
Thus, the correct answer is A .
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