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Geometric Integers (Posted on 2006-11-09) Difficulty: 2 of 5
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?

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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doing something wrong? | Comment 1 of 6

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 2006-11-09 11:04:09
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