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

(New page: ==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...)
 
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==Problem==
 
==Problem==
  
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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: <math>\boxed{1:01}</math>, <math>\boxed{4:44}</math>, and <math>\boxed{12:21}</math>.  How many times during a <math>12</math>-hour period will be palindromes?
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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: <math>\boxed{1:01}</math>, <math>\boxed{4:44}</math>, and <math>\boxed{12:21}</math>.  How many times during a <math>12</math>-hour period will be palindromes?
  
 
<math>\text{(A)}\ 57 \qquad \text{(B)}\ 60 \qquad \text{(C)}\ 63 \qquad \text{(D)}\ 90 \qquad \text{(E)}\ 93</math>
 
<math>\text{(A)}\ 57 \qquad \text{(B)}\ 60 \qquad \text{(C)}\ 63 \qquad \text{(D)}\ 90 \qquad \text{(E)}\ 93</math>
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==See Also==
 
==See Also==
  
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[[1988 AJHSME Problems]]
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{{AJHSME box|year=1988|num-b=24|after=Last<br>Problem}}
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]

Revision as of 17:07, 4 June 2009

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 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last
Problem
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All AJHSME/AMC 8 Problems and Solutions
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