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Topic: Digital Transference Problem Revisited (Read 746 times) 

K Sengupta
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Digital Transference Problem Revisited
« on: Nov 28^{th}, 2005, 12:29am » 
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Considering a K digit integer N ( where the last digit of N is nonzero) the number L is constituted by deleting the first P digits of N and shifting a permutation of these P digits ( the definition of the said permutation being inclusive of the original first P Digits of N) to the end of N. IF: (i) L is divisible by N such that L is not equal to N, determine the total number of pairs (G,N) where 2<=P<=6 ,8<=K<=15 and Max(L,N)<10^15 (ii) If, in addition, the sum of the digits in N is a perfect Mth power with M being a positive whole number grater than 1, determine the total number of distinct Quadruplets ( P,K, L,N) where 2<=P<=6 ,8<=K<=15 and Max(L,N)<10^15. (iii) Determine the minimum possible magnitude of the pair (L,N) separately for each pair (P,K) where 2<=P<=6 ,8<=K<=15 and Max(L,N)<10^15.


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Grimbal
wu::riddles Moderator Uberpuzzler
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Re: Digital Transference Problem Revisited
« Reply #1 on: Nov 28^{th}, 2005, 2:38am » 
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I don't even understand the question. What are you doing to N? Can you give an example?


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K Sengupta
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Re: Digital Transference Problem Revisited
« Reply #2 on: Nov 30^{th}, 2005, 11:59pm » 
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Suppose N=32456789and for example we consider the first 3 digits(P=3,K=8) ,i.e.,352. All possible permutation of 352(including itself) are 324,342,243,234,423,432 so that this gives six available values of L by which are 56789324, 56789342,56789243,56789234,56789423,56789432. Clearly, all these six values of L may or may not satisfy all the three conditions. In case of multiple L values for a single N corresponding to a given choice of (P,K) , satisfying conditions of the problem  for example if there are 3 values of L La1,La2 and La3 say for a single N=Na,P=Pa and K=Ka then we would have 3 distinct quadruplets for (N,P,K,L) corresponding to N=Na,P=Pa and K=Ka given by (N,P,K,L)= (Na,Pa,Ka,La1),(Na,Pa,Ka,La2) and (Na,Pa,Ka,La3).

« Last Edit: Dec 1^{st}, 2005, 3:30pm by Icarus » 
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