# 1988 AJHSME Problems/Problem 25

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## 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.

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