Mathematics-Online problems: Solution to the problem of the (previous) week

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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

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 respectively appear each for ten hours $(10 \cdot 60 = 600\,$ times and the digit 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:

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

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