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?
(In reply to
computer solution by Charlie)
Good find, Charlie! Not sure how I missed that, it seems so simple...
Still not sure why it gave me such a headache the first time around. We are looking for three integers of the form a, ax, and ax^2. These can be combined in three ways:
a(ax) + ax^2 = 201
a(ax^2) + ax = 201
ax(ax^2) + a = 201
Of those three it can be shown that only the last would work, since the lefthand side of the first two equations will always produce an even number.
So a is our remainder, and it must be odd. Since ax and ax^2 are integers, x is some fraction h/k where h and k are integers and both k and k^2 divide a.
Ignoring the case where a = k = 1, the first possible value of a is 9, and k is 3. Plugging these values in we get:
9 + 3h^3 = 201
3h^3 = 192
h^3 = 64
h = 4
So then x is 4/3, and our three values are:
Remainder = a = 9
Divisor = ax = 9*4/3 = 12
Quotient = ax^2 = 9*16/9 = 16

Posted by tomarken
on 20061109 15:50:51 