Question 1207683
<pre>
Possible remainders are integers from 0 through 16, inclusive: 

  P =  Q x 17 +  R
------------------
 17 =  1 x 17 +  0
103 =  6 x 17 +  1
  2 =  0 x 17 +  2
  3 =  0 x 17 +  3
 89 =  5 x 17 +  4
  5 =  0 x 17 +  5
193 = 11 x 17 +  6
  7 =  0 x 17 +  7
 59 =  3 x 17 +  8
179 = 10 x 17 +  9
 61 =  3 x 17 + 10
 11 =  0 x 17 + 11
 29 =  1 x 17 + 12
 13 =  0 x 17 + 13
 31 =  1 x 17 + 14
 83 =  4 x 17 + 15
 67 =  3 x 17 + 16

This might make you wonder if this is true:

For any prime t and any integer r, 0 <u><</u> r <u><</u> t-1,
there will always be a prime p such that when 
p is divided by t, the remainder will be r.

We just proved it true for t=17.  

Edwin</pre>