Three
distinct 3-digit positive decimal (base 10) integers
P,
Q and
R, having no leading zeroes and with
P >
Q >
R, are such that:
(i)
P,
Q and
R (in this order) are in geometric sequence, and:
(ii)
P,
Q and
R are obtained from one another by
cyclic permutation of digits.
Find all possible triplet(s)
(P, Q, R) that satisfy the given conditions.
Note: While the solution may be trivial with the aid of a computer program, show how to derive it without one.
Observations:
1) Let P = 100a + 10b + c where a,b,c are integers.
In order to be cyclic where P > Q > R.
a>= b > c
For instance 441 would do.
Because P, Q and R are geometric, however, I expect that
a > b > c
2) P = Q = R (mod 9) (by casting out 9's)
Because the numbers are geometric, this is most easy if
P = Q = R = 0 (mod 9)
In that case, a + b + c = 9 or 18
The only possibilities are 432, 531, 621, 765, 873, 864,
981, 972, 963, and 954.
Only two of these work:
432 > 324 > 243 and
864 > 648 > 486
3) We can rule out the case where P = Q = R <> 0 (mod 9),
because this would require P/Q = 1 mod 9.
But P/Q = 1 makes all numbers the same,
and P/Q >= 10 makes P/R >= 100, which is not possible.
So, I think the only values for P are 432 and 864.
Solution? (spoiler?)
