Question 1193059
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<pre>

Of 4 possible digit positions, the first position is just occupied by the digit of 5.


Three other positions are free, and we can put any of 10 possible digits from 0 to 9 in any of these 3 positions.


It gives  {{{10^3}}} = 1000 possible digit codes.    <U>ANSWER</U>
</pre>

Solved (correctly).


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Ignore answer by @Theo, since it is INCORRECT.


I don't know why @Theo decided about 9 digits.


Working base 10, we always have 10 possible digits.