Decimal integers with nonzero digits and sum of powers puzzle

In summary, there are two possible positive decimal integers, P = 49 and P = 79, that satisfy the equation P = X1^X1 + X2^X2 + ... + Xn^Xn, where P >= 2 and none of the digits in P are zero. This is assuming that X1, X2, X3, etc. are all individual digits of P and that 0^0 = 1. The first solution, P = 49, is the only one that meets the additional condition that all digits in P must be distinct. The upper bound for possible solutions is estimated to be around 3.4 billion.
  • #1
K Sengupta
113
0
Determine all possible positive decimal integer(s) P = X1X2X3….Xn, where P>=2 with none of the digits in P being zero, that satisfy this equation:

P = X1^X1 + X2^X2 + ……+ Xn^Xn

(For example, P = 234 cannot be a solution since 2^2 + 3^3 + 4^4 is equal to 287, not 234.)

Notes:

(i) X1X2X3….Xn denotes the concatenation of the digits X1, X2, …, Xn and do not represent the product of the digits.

(ii) P cannot admit any leading zero.
 
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  • #2
I'm assuming that X1, X2, X3 are all the individual *digits* of P?

DaveE
 
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  • #3
davee123 said:
I'm assuming that X1, X2, X3 are all the individual *digits* of P?

DaveE

Yes, each of X1,X2, ...,Xn correspond to an individual digit of P.
 
  • #4
Can any digit be the same as another digit?
 
  • #5
K Sengupta said:
(ii) P cannot admit any leading zero.
...or presumably, any other zeros.
 
  • #6
Here are a few:
1
3435
438579088
Of these only the first meets the unstated condition that all digits be distinct. The third one fails to meet the unstated condition that no digits be 0. I used a brute force method to get this partial solution. I do not know if this list is complete.
 
  • #7
That also assumes that 0^0 = 0-- I thought the more accepted solution was that 0^0 = 1?

Otherwise, I think there's an upper bound of roughly 3.4 billion. Beyond that I think the rate at which the sum of the powers increases is capped (since 9^9 is the highest sum a digit can contribute), and the number itself is increasing more quickly.

DaveE
 
  • #8
davee123 said:
That also assumes that 0^0 = 0-- I thought the more accepted solution was that 0^0 = 1?

Good catch. There was a bug in my code. So I have only found 2 solutions.
 

FAQ: Decimal integers with nonzero digits and sum of powers puzzle

1. What is the "Decimal Integers with Nonzero Digits and Sum of Powers Puzzle"?

The "Decimal Integers with Nonzero Digits and Sum of Powers Puzzle" is a mathematical puzzle that involves finding a set of integers that satisfies a specific criteria. The criteria is that each integer in the set must contain only nonzero digits and the sum of their powers must equal a given number.

2. How do you solve the "Decimal Integers with Nonzero Digits and Sum of Powers Puzzle"?

The puzzle can be solved by using trial and error or by using a mathematical approach. The mathematical approach involves breaking down the given number into its prime factors and then finding combinations of those factors that can be used to create the desired sum of powers.

3. Are there any specific rules or restrictions for solving this puzzle?

Yes, there are a few rules that must be followed when solving this puzzle. Firstly, each integer in the set must contain only nonzero digits. Additionally, the sum of the powers of the digits in each integer must equal the given number. Lastly, each integer can only be used once in the set.

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Although this puzzle may seem like a purely theoretical exercise, it actually has practical applications in fields such as cryptography and coding theory. It can also be used as a fun and challenging brain teaser.

5. Is there a specific strategy or approach that can be used to solve this puzzle?

There is no single strategy that works for every instance of this puzzle. However, some common approaches include breaking down the given number into its prime factors, starting with smaller numbers and working up to larger ones, and using a combination of trial and error and mathematical reasoning. The best strategy may vary depending on the specific puzzle and the solver's personal preferences.

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