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Lychrel


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#1 BMAD

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Posted 08 August 2013 - 01:45 AM

If we take 47, reverse and add, 47 + 74 = 121, which is palindromic. Not all numbers produce palindromes so quickly.

For example, 349 + 943 = 1292, 1292 + 2921 = 4213 4213 + 3124 = 7337

That is, 349 took three iterations to arrive at a palindrome. Although no one has proved it yet, it is thought that some numbers, like 196, never produce a palindrome. A number that never forms a palindrome through the reverse and add process is called a Lychrel number.

Due to the theoretical nature of these numbers, and for the purpose of this problem, we shall assume that a number is Lychrel until proven otherwise.

In addition you are given that for every number below ten-thousand, it will either (i) become a palindrome in less than fifty iterations, or, (ii) no one, with all the computing power that exists, has managed so far to map it to a palindrome. In fact, 10677 is the first number to be shown to require over fifty iterations before producing a palindrome: 4668731596684224866951378664 (53 iterations, 28-digits).

Surprisingly, there are palindromic numbers that are themselves Lychrel numbers; the first example is 4994. How many Lychrel numbers are there below ten-thousand?

(Here's a problem for my programmers)
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#2 Anza Power

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Posted 09 August 2013 - 07:46 AM

Code:

 

#include <iostream>
#define BUF_SIZE 100
using namespace std;
 
void putInArray(int num, int* buf){
    for(int i=0;i<BUF_SIZE;i++){
        buf[i] = num % 10;
        num = num / 10;
    }
}
 
bool isPalindrom(int* buf){
    int i=0, j=BUF_SIZE-1;
    while(buf[j]==0 && j>0){
        j--;
    }
    while(i<j){
        if(buf[i] != buf[j]){
            return false;
        }
        i++;
        j--;
    }
    return true;
}
 
void applyIteration(int* buf){
    int i=0, j=BUF_SIZE-1;
    while(buf[j]==0 && j>0){
        j--;
    }
    while(i<=j){
        int temp = buf[i] + buf[j];
        buf[i++] = temp;
        buf[j--] = temp;
    }
    
    for(int i=0;i<BUF_SIZE-1;i++){
        buf[i+1] += buf[i] / 10;
        buf[i] = buf[i] % 10;
    }
}
 
bool isLychrel(int* buf){
    for(int i=1;i<50;i++){
        applyIteration(buf);
        if(isPalindrom(buf)){
            return false;
        }
    }
    return true;
}
 
int main() {
    int buf[BUF_SIZE];
    
    for(int num=0;num<10000;num++){
        putInArray(num,buf);
        if(isLychrel(buf)){
            cout << num << " is Lychrel " << endl;
        }
    }
}

Edited by Anza Power, 09 August 2013 - 07:52 AM.

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#3 Anza Power

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Posted 09 August 2013 - 07:55 AM

Spoiler for code

 
Spoiler for result

 
Also: http://projecteuler.net/problem=55

Edited by Anza Power, 09 August 2013 - 07:57 AM.

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