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        riddles >> medium >> digit powers: 2 other cases
        
(Message started by: Christine on Feb 22^nd, 2012, 2:10pm)

Title: digit powers: 2 other cases
Post by Christine on Feb 22^nd, 2012, 2:10pm

Case #1

1364 & 6143
(1^6) + (3^1) + (6^4) + (4^3) = 1364

where each digit of 1364 is raised to digits of 6143 yielding to the first number 1364.
The numbers 1364 and 6143 contain the same digits.
Here's another example,

4355 & 5435
(4^5) + (3^4) + (5^3) + (5^5) = 4 355

And,
067236 & 326706
(3^0) + (2^6) + (6^7) + (7^2) + (0^3) + (6^6) = 326 706

How often does it happen? And also, in the case of a leading digit 0? How to find others?

Case 2

4096 = 4^6 + 0^9
Notice the number 4096 is sum of digits raised to power of digits, all these of digits are in this number.

Other examples,
397612 = 3^2 + 9^1 + 7^6 + 6^7+ 1^9+ 2^3
48625 = 4^5 + 8^2 + 6^6 + 2^8 + 5^4

How to find others? Is there a formula to find them?

Title: Re: digit powers: 2 other cases
Post by Michael Dagg on Feb 27^th, 2012, 2:10pm

I found that some more numbers less than 50 000 which have your property
(in base 10) are 1364, 3435, 4155, 4316, 4355, 15630, 17463, 48625, 38650 .
The list is definitely finite! If the property you created were to hold
for an n-digit number N , then N would have to be the sum of n numbers,
each no larger than 9^9 (or (B-1)^(B-1) if you are writing in base B),
so the sum would be at most n(9^9) < n(10^9) , whereas the number is at
least 10^(n-1) . So the sum of the digit powers is too small, unless
n(10^9) >= 10^(n-1) , i.e. unless n >= 10^(n-10) . But this is certainly impossible
if n is 12 or more. So your property will never be satisfied by any
12-or-more-digit number.