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        riddles >> hard >> 15 DIGIT NUMBERS
        
(Message started by: K_Sengupta on Nov 21^st, 2005, 12:11am)

Title: 15 DIGIT NUMBERS
Post by K_Sengupta on Nov 21^st, 2005, 12:11am

Let R(X) be the number constituted by reversing the digits of X.

(i) When a 15-digit positive Palindromic integer P is divided by a 12 -digit positive whole number N, a remainder of R(N) is obtained.

(ii) Dividing N by R(N) yields a remainder M, where M is a prime number. The last digit of neither P nor N is zero.

Determine the total number of quadruplets
(P,N, R(N), M) satisfying these conditions.

If in addition, the sum of the digits of M is a perfect square what would be the total number of quadruplets satisfying conditions of the problem?

If however sum of the digits of M was a perfect cube instead of perfect square would there be any solution to this problem? If the answer is in the affirmative, what would have been the total number of quadruplets in that situation?

Title: Re: 15 DIGIT NUMBERS
Post by K_Sengupta on Nov 25^th, 2005, 11:05pm

I have modified this problem by adding another condition.