Self obsessed numbers

Computer Science Level 4

Consider a n n digit integer x = a n a n − 1 ⋯ a 1 ‾ x=\overline{a_na_{n-1}\cdots a_1} .

Let us call the number x x to be self obsessed if

x = a n a n − 1 ⋯ a 1 ‾ = a n a n + a n − 1 a n − 1 + ⋯ + a 1 a 1 x=\overline{a_na_{n-1}\cdots a_1}=a_n^{a_n}+a_{n-1}^{a_{n-1}}+\cdots+a_1^{a_1} .

It is easy to see that the smallest such integer is x = 1 x=1 .

What is the next integer with this property?

HINT : 0 0 0^0 is . Hence, the answer cannot have zero as one of its digits.


The answer is 3435.

Janardhanan Sivaramakrishnan by Janardhanan Sivaramakris… , 42

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6 solutions

Samarpit Swain
Aug 28, 2015

C++ 

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#include <iostream>
#include<math.h>
using namespace std;

int main() {
unsigned long long int n=2,num,dig,sum;
 do
   {    num=n;
      sum=0;
  while(num>0)
{    dig=num%10;
         if(dig==0)
        break;
  sum+=pow(dig,dig);
    num/=10;
    }
  n++;
          }while(n!=sum)
    cout<<n<<endl;
return 0;
}

Output: 

1
 
3435

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Arulx Z
Aug 27, 2015 
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>>> count = 2
>>> while count != sum(int(x) ** int(x) for x in str(count)):
        count += 1
>>> print count
3435

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Simple standard approach.

3 3 + 4 4 + 3 3 + 5 5 = 27 + 256 + 27 + 3125 = 3435 3^3+4^4+3^3+5^5=27+256+27+3125=\boxed{3435}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Just a cool fact: these are called Münchhausen numbers! The next one, courtesy of OEIS, is 438579088.

Ryan Tamburrino - 6 years, 4 months ago

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That is assuming 0 0 = 0 0^0=0 if you read the hint in the question, it eliminates 438579088.

Janardhanan Sivaramakrishnan - 6 years, 4 months ago
David Holcer
Nov 7, 2015 
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c=2
while True:
   summ=0
   for i in str(c):
       summ+=(int(i)**int(i))
   if (summ==c):
       print c
       break
   c+=1

Simple enough approach, using string manipulation and breaking out of infinite loops.

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Bill Bell
Oct 5, 2015

The hint is useful. You can also eliminate some wasted time by noting that there is no point in adding terms once the sum is at least as great as x x . 

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def isSelfObsessed(m):
    n=m
    r = 0
    while n:
        d=n%10
        if d==0:
            return False
        r=r+d**d
        if r>m:
            return False
        n=n/10
    return r==m

m=1
while True:
    m+=1
    result=isSelfObsessed(m)
    if result:
        print m
        break

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Prakhar Gupta
Feb 3, 2015

Here is a J A V A JAVA code for this problem.

It must be a computer Science problem.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

But solvable using the pencil & paper approach as well. I did it that way, so it's not a rigid cs problem.

Saya Suka - 1 year, 2 months ago

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