20200621, 05:20  #1 
Sep 2018
2×3 Posts 
Münchhausen numbers of length 3
3435 is wellknown as the Münchhausen number in the base 10 and its length (number of digits) is 4.
Are there infinitely many Münchhausen numbers of length 3? Please visit http://searchfornumbers.blogspot.com...length3.html 
20200621, 06:04  #2 
Jun 2003
2^{2}×3^{2}×131 Posts 
904:[1, 7, 1]:823545

20200621, 15:39  #3 
Sep 2018
2×3 Posts 
Great!

20200621, 16:19  #4 
Jan 2017
79 Posts 
No larger solutions with less than 1000 base10 digits. Given the three digits, you can solve for base as a seconddegree polynomial. I checked that all triples with largest digit in [8, 400[ give irrational bases, and 400^400 would have 1041 base10 digits.

20200621, 20:53  #5 
"Dylan"
Mar 2017
3×173 Posts 
So I wrote some code in Python to test this:
Code:
#Python script to find Munchhausen numbers for some base and length #A Munchhausen Number is a number equal to the sum of it's digits #raised to each digit's power. #For example, in base 10 3435 is a Munchhausen number, since #3435 = 3*10^3+4*10^2+3*10+5 and 3435 = 3^3+4^4+3^3+5^5=27+27+256+3125. #The question posed here: https://mersenneforum.org/showpost.php?p=548680&postcount=1 #are there infinitely many Munchhausen numbers with length 3? #first, we define a function for the sum of the digits raised to each digit's power. def sum_digits_raised(digits): '''Takes a list of integers and sums up the powers of them.''' digitpowersum = 0 for digit in digits: if digit < 0: raise ValueError("Digit needs to be nonnegative.") digitpowersum += pow(digit, digit) #print(digitpowersum) return digitpowersum #as a check, run the list [3,4,3,5]. It should yield 3435 #print(sum_digits_raised([3,4,3,5])) #Now, we define a function to translate from base n to base 10. def tobase10(orig_base, digits): '''Converts a list of digits in base orig_base into a number in base 10.''' base10num = 0 i = len(digits)1 for digit in digits: base10num += digit*pow(orig_base, i) i = 1 return base10num #check on [3,4,3,5] in base 10 #print(tobase10(10, [3,4,3,5])) #Now we want to run for length 3 and base all the possible digit combinations. base = 904 biggestlefthandside = tobase10(base, [base1,base1,base1]) #this works for length 3, the problem at hand. for i in range(0, base+1): for j in range(0, base+1): for k in range(0, base+1): digits = [i,j,k] righthandside = sum_digits_raised(digits) #the biggest we could hope the left hand side to be is #(base1)*base^(length1)+(base1)*base^(length2)+...+base1 #so if the right hand side is bigger than that there's no way the lefthand side will be the same as the right. if righthandside > biggestlefthandside: continue lefthandside = tobase10(base, digits) if lefthandside == righthandside: print(i, j, k, str(tobase10(base, digits))) print("Incrementing i to " + str(i+1)) 
20200621, 22:56  #6 
"Robert Gerbicz"
Oct 2005
Hungary
17×83 Posts 
A faster approach: if the discriminant is not a square then you can eliminate that triplet. With that you don't even need to calculate the coefficients, just check for ~50 smallest primes if the discriminant is a quadratic residue or not. Very likely there will be no larger solution.

20200621, 23:21  #7  
Jan 2017
79 Posts 
Quote:
Code:
base = 904 p = [i**i for i in range(base)] d = {pii:i for i, pi in enumerate(p)} for i in range(base): ni = base*base*i  p[i] for j in range(base): if ni + base*j  p[j] in d: print(i, j, d[ni+base*jp[j]]) 

20200621, 23:26  #8  
Jan 2017
4F_{16} Posts 
Quote:


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