<PRE><B>Question 1196099</B>
&lt;br&gt;
The problem as posted has a large number of solutions.  I suspect the information is not shown correctly.&lt;br&gt;
(1) a=b+2 and b=c+2 together mean the first three digits of the number are one of these:&lt;br&gt;
420, 531, 642, 753, 864, or 975&lt;br&gt;
That&#39;s 6 choices for the first three digits.&lt;br&gt;
(2) d=2d/2 tells us nothing; e=2d/4=d/2d tells us that d is even.  So d can be any even digit and e is half of d.  So the last two digits are one of these:&lt;br&gt;
00, 21, 42, 63, or 84&lt;br&gt;
That&#39;s 5 choices for the last two digits.&lt;br&gt;
With the problem as shown, there are 6*5=30 5-digit numbers that satisfy the given conditions.&lt;br&gt;

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