Difference between revisions of "1988 AJHSME Problems/Problem 25"

m
 
Line 17: Line 17:
 
{{AJHSME box|year=1988|num-b=24|after=Last<br>Problem}}
 
{{AJHSME box|year=1988|num-b=24|after=Last<br>Problem}}
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
 +
{{MAA Notice}}

Latest revision as of 23:56, 4 July 2013

Problem

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: $\boxed{1:01}$, $\boxed{4:44}$, and $\boxed{12:21}$. How many times during a $12$-hour period will be palindromes?

$\text{(A)}\ 57 \qquad \text{(B)}\ 60 \qquad \text{(C)}\ 63 \qquad \text{(D)}\ 90 \qquad \text{(E)}\ 93$

Solution

From $1$ to $9$, the times will be of the form $\boxed{a:ba}$. There are $9$ choices for $a$ and $6$ choices for $b$, so there are $9\cdot 6 =54$ times in this period.

From $10$ to $12$, the minutes are already determined, so there are only $3$ times in this case.

In total, there are $54+3=57\rightarrow \boxed{\text{A}}$ palindromic times.

See Also

1988 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last
Problem
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png