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Topic: 15 DIGIT NUMBERS (Read 299 times) |
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K Sengupta
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15 DIGIT NUMBERS
« on: Nov 21st, 2005, 12:11am » |
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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?
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| « Last Edit: Nov 25th, 2005, 11:03pm by K Sengupta » |
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K Sengupta
Senior Riddler
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Re: 15 DIGIT NUMBERS
« Reply #1 on: Nov 25th, 2005, 11:05pm » |
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I have modified this problem by adding another condition.
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