# Lychrel Numbers

[FINAL SOLUTION] Lychrel Numbers

The previous solution was already almost correct. The only change we need to do is in the is_lychrel method. We change this (wrong) code:

```def is_lychrel(number, invocation):
if invocation > MAX_ITERATIONS:
return True

if is_palindrome(number):
return False

return is_lychrel(number + int(reverse_number(number)), invocation + 1)
```

to

```def is_lychrel(number, invocation):
if invocation > MAX_ITERATIONS:
return True

result = number + int(reverse_number(number))

if is_palindrome(result):
return False

return is_lychrel(result, invocation + 1)
```

The change really is that we no longer check is_palindrome on the input number but rather on the result of the calculation.

The complete and correct solution looks like this:

```#!/usr/bin/env python

MAX_ITERATIONS = 50
NRS_TO_CHECK = 10000

def reverse_number(number):
return str(number)[::-1]

def is_palindrome(number):
nr_as_string = str(number)
reverse_as_string = reverse_number(number)
return str(number) == reverse_as_string

def is_lychrel(number, invocation):
if invocation > MAX_ITERATIONS:
return True

result = number + int(reverse_number(number))

if is_palindrome(result):
return False

return is_lychrel(result, invocation + 1)

lychrel_numbers = 0
for cur_number in range(1, NRS_TO_CHECK):
if is_lychrel(cur_number, 0):
lychrel_numbers += 1
if is_palindrome(cur_number):
print 'is lychrel and initial palindrome: ' + str(cur_number)

print 'Lychrel numbers below {0}: {1}'.format(NRS_TO_CHECK, lychrel_numbers)
```

Running this code now returns the correct result

```Lychrel numbers below 10000: 249
```

If you made it this far, well done and thank you for sticking around. Feel free to leave any questions by leaving a comment and we will get back to you.

Lynx