A^5+B^5+C^5+D^5+E^5=S
Given that the concatenation ABCDE digits is a 9-digit zero-less 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 10-digit 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 1-digit while the remaining four are 2-digits.
Summing any set of four 2-digit numbers and one 1-digit number where each digit is unique and the total is smaller than a 10-digit number, the highest 2-digit number is 59. The highest 9-digit 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.
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Posted by Dej Mar
on 2015-03-20 00:58:55 |
re: computer solution
