Getting Cubic With T^2 (Posted on 2007-10-10)

	Submitted by K Sengupta
Rating: 5.0000 (1 votes)
Solution:	(Hide)
T = 1, 27 are the only possible values satisfying the conditions of the problem. EXPLANATION: Let S(n) denote the sum of digits of n, then by the problem we have : T^2 = S(T)^3…..(i) Now, there must exist some non negative integer p, such that: 10^p <= T < 10^(p+1), so that: 10^(2p) <= T^2 = S(T)^3 < = (9p)^3 .....(*) Since, 100*(p/(p+1))^3 >= 100*(2/3)^3 = 29.629......, for p>=2, and (p/(p+1))^3   is strictly increasing in p for positive integers p . Accordingly, it follows by induction that 10^(2p) > (9p)^3, whenever p >=2, so that the relationship (*) is only satisfied for non negative integers p < 2 , so that T < 10^2 = 100. Since T<100, it follows that S(T) <= (9+9) = 18. Now, T^2 = S(T)^3, so that T must be a perfect cube as well as a perfect square. Hence, S(T) is a perfect square. The only possibilities with S(T)  < = 18 are S(T) = 1, 4, 9, 16, giving T = 1, 8, 27, 64 in terms of (i)......(#) For T =8, we have S(T) = 8!= 4, a contradiction. Similarly, for T = 64, we have S(T) =10 !=16, a contradiction. For T = 1 and 27 , the respective values of S(T) are 1 and 9 which are in conformity with (#). Consequently, T = 1, 27 are the only possible values satisfying the conditions of the problem.

	Subject	Author	Date
	re: Solution?	badger	2007-10-10 14:53:14
	re: computer analysis/solution	Dej Mar	2007-10-10 11:44:31
	computer analysis/solution	Charlie	2007-10-10 10:44:56
	Solution?	Fernando	2007-10-10 10:32:15