objective c - Find the 42th number which sum of its digits is 42 -

i've got interesing programming/mathematical problem. want write application, find 42th number sum of digits 42, need find number have sum of digits 42 , add new array print 42th number in new array have 50 numbers have sum of digits 42. thank taking time help.

@implementation appdelegate

static nsarray *newarray ; - (void)applicationdidfinishlaunching:(nsnotification *)anotification {       int numberwithsum42;     (int i=69999;i<80000;i++)     {         int x=[self findtotalnumber:i];          if(x==12){              numberwithsum42=i;             newarray = [nsarray arraywithobjects:[nsnumber numberwithint:numberwithsum42], nil];         }         nslog(@"%@",newarray);     }     [self print42ndvariable]; } -(void)print42ndvariable{     int j;     int count = (int)[newarray count];     (j = 0; j < count; j++){         nslog (@"42nd variable of array   = %@", [newarray objectatindex: 42]);     }  } -(int)findtotalnumber:(int) n{     int s=0;     while (n>0)     {         int k=n%10;         s=s+k;         n=n/10;     }     nslog(@"%i",s);     return n; } 

@end

- (void)applicationdidfinishlaunching:(nsnotification *)anotification{ int requirednumber; int currentindex; int sumofdigits;  //assuming limits wnat 69000<=x<80000  (int i=69999;i<80000;i++) {     sumofdigits= [self sum:i];      if(i==42){          currentindex++;     }     if (currentindex == 42){          nslog(i); //the number!     } }  } -(void)sum:(int)number{ int digit,sum=0, temp;     temp = number;      while(temp > 0) {         digit = temp%10;         sum += digit;         temp = temp/10;     }     nslog(@"sum of digits of %i = %i",number,sum);     return sum; }