Curious Cyclic Circumstance (Posted on 2009-02-07)

	Submitted by K Sengupta
Rating: 5.0000 (1 votes)
Solution:	(Hide)
(P, Q, R) = (983, 839, 398) is the only triplet that satisfies all the given conditions. EXPLANATION: Denoting sod(n) as the sum of digits of n, we observe that: sod(P*Q*R) = 8+6+5+4+2*3+3*2 = 35, so that: sod(P*Q*R) (mod 9) = 8 Since P, Q and R are cyclic permutations of each other, we must have: sod(P) = sod(Q) = sod(R). Let the common value be M. Then, M^3 (mod 9) = 8, so that: M (mod 9) = 2, 5 or 8. Since P, Q and R are cyclic permutations of each other, it follows that each of P, Q and R is constituted by the same three digits (x,y and z, say) but in different order. Since P > Q> R, it follows that at least two of x,y and z must be different. Without loss of generality, we can set x ≤ y ≤ z, so that: xyz = R. Now, the minimum value of P*Q*R is 222334568. But, we observe that: (199)*(919)*(991) = 181235071 < 222334568, and hence, x ≥ 2. Accordingly, we have: If M(mod 9) = 2, then x+y+z = 11, or 20 If M(mod 9) = 5, then x+y+z = 14, or 23 If M(mod 9) = 8, then x+y+z = 8, 17, or 26 Since the last digit of the product P*Q*R is 6, it follows that the last digit of x*y*z is also 6. We now observe that the valid triplets in conformity with the given conditions are: (x,y,z) = (2,2,4), (2,3,6), (3,8,9), (4,4,6) and, (4,8,8). If (x,y,z) = (2,2,4), then R = 224, Q= 242, and P = 422. But, the product P*Q*R = 22875776 has only 8 digits. Similarly, for (x,y,z) = (2,3,6), (4,4,6) and (4,8,8) the none of the products product P*Q*R does not conform to the given structure. For (x,y,z) = (3,8,9), we have: (P, Q, R) = (983, 839, 398), whereby: P*Q*R = 328245326, which is in consonance with the given structure. Consequently, the required triplet in conformity with the given conditions is: (983, 839, 398).

	Subject	Author	Date
	computer solution	Charlie	2009-02-07 17:00:51