The number 201 is divided by a positive integer N. It is observed that the quotient, remainder and divisor (that is, N itself), but not necessarily in this order, are in geometric sequence.
What can N be?
q=quotient, d=divisor, r=remainder, x=ratio of geometric sequence.
q*d + r = 201
There are six cases to consider. Because the terms can be in any order.
Case 1: Consider q<d<r then d=qx and r=qx^2
q*qx + qx^2 = 201
q(qx + x^2) = 201
Now 201 can only be a product as 3*67 or 1*201 so q must be one of these 4 values.
3x + x^2 = 67 has no rational solutions
67x + x^2 = 3 has no rational solutions
x^2 + x =201 has no rational solutions
x^2 + 201x = 1 has no rational solutions
Proceeding through cases 2 through 6 yields the same results.
Does this prove there are no solutions or is something wrong with my reasoning?

Posted by Jer
on 20061109 11:04:09 