Question 1184385
<pre>
Let the whole number with the digit 4 removed be N.

Then the whole number = 10N+4

4 times this whole number = 4(10N+4) = 40N+16

This must equal to p = N + 4*10^a, where a is some integer. So

{{{40N+16 = N+4*10^a}}}

{{{39N+16 = 4*10^a}}}

The right side is 40000000...0, but we don't know what "a" is so
we don't know where the 0's stop.

However we do know that 

dividend = quotient*divisor + remainder

 4*10^a  =        N*39      +    16

40000... =        N*39      +    16

So we start dividing 39 into 40000000... and if we ever get a remainder
of 16, we will have our answer for N  So let's see if we can get a 16
remainder:
            <u> 10256      </u>
         39)400000000000...
            <u>39</u>
             10
             <u>00</u>
             100
              <u>78</u>
              220
              <u>195</u>
               250
               <u>234</u>
                16    <--aha! we got a remainder of 16.

So N, the number without the 4 on the right end is 10256

So M = 102564 and p = 4M = 410256

Edwin</pre>