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Topic: digit powers: 2 other cases (Read 1445 times) |
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Christine
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Posts: 159
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digit powers: 2 other cases
« on: Feb 22nd, 2012, 2:10pm » |
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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?
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Michael Dagg
Senior Riddler
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Posts: 500
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Re: digit powers: 2 other cases
« Reply #1 on: Feb 27th, 2012, 2:10pm » |
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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.
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| « Last Edit: Feb 27th, 2012, 2:16pm by Michael Dagg » |
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Regards,
Michael Dagg
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