A^5+B^5+C^5+D^5+E^5=S
Given that the concatenation ABCDE digits is a 9digit zeroless pandigital number
and so is the sum S  there is only one sum S to make the above equation true.
Find it.
(In reply to
computer solution by Charlie)
As x^5, such that x is a positive integer composed of distinct digits and is less than a 10digit number, the highest value of x is 62. Thus, the five numbers, A, B, C, D, and E, are each two or less digits in length. Given that ABCDE is 9 digits, one of the five partitioned numbers is 1digit while the remaining four are 2digits.
Summing any set of four 2digit numbers and one 1digit number where each digit is unique and the total is smaller than a 10digit number, the highest 2digit number is 59. The highest 9digit number thus composed is 995702359.
The list of possible arrangements fed as a file into a computer program can be reduced by eliminating any number greater than 59.

Posted by Dej Mar
on 20150320 00:58:55 