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Mathematics-Online problems:

Solution to the problem of the (previous) week


Problem:

During a day, the display of a digital watch shows $ 24 \cdot 60$ different times consisting of $ 4$ digits each, that is $ 4 \cdot 24 \cdot 60 = 5760$ digits altogether:

from     \includegraphics[width=0.1\linewidth]{TdM_11_A4_bild1}     to      \includegraphics[width=0.1\linewidth]{TdM_11_A4_bild2}

How many times does each of the digits $ 0, 1,\, \ldots\, , 9$ occur?


Answer:

\includegraphics[width=2ex]{TdM_11_A4_bild_digit_0} :            \includegraphics[width=2ex]{TdM_11_A4_bild_digit_1} :  
                   
\includegraphics[width=2ex]{TdM_11_A4_bild_digit_2} :       \includegraphics[width=2ex]{TdM_11_A4_bild_digit_3} :  
                   
\includegraphics[width=2ex]{TdM_11_A4_bild_digit_4} :       \includegraphics[width=2ex]{TdM_11_A4_bild_digit_5} :  
                   
\includegraphics[width=2ex]{TdM_11_A4_bild_digit_6} :       \includegraphics[width=2ex]{TdM_11_A4_bild_digit_7} :  
                   
\includegraphics[width=2ex]{TdM_11_A4_bild_digit_8} :       \includegraphics[width=2ex]{TdM_11_A4_bild_digit_9} :  


Solution:

We regard the four digits on the display separately.

First digit:
The digits 0 and $ 1$ respectively appear each for ten hours $ (10
\cdot 60 = 600\,$times$ )$ and the digit $ 2$ for four hours $ (240\,$times$ ).$

Second digit:
Each of the digits $ 0,\, 1,\, 2,\, 3$ are to be seen for three hours $ (180\,$times$ ).$ All the other digits do not appear between 20:00 and 23:59. They are displayed only $ 120\,$times$ .$

Third digit:
Here, only the digits $ 0,\, \ldots , 5$ appear, and all of these for the same number of times, that is $ 24 \cdot 60 / 6 =
240\,$times$ .$

Fourth digit:
All the digits appear $ 24 \cdot 6 = 144\,$times$ .$


To sum up we can compile the following table:

\begin{tabular}{r\vert r\vert r\vert r\vert r\vert r\vert r\vert r\vert r\vert r...
... 504} & {\bf 264} & {\bf 264} & {\bf 264}
& {\bf 264} & \\
\par
\end{tabular}


[problem of the week]