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?

NOTE: Wording was modified slightly on 24 April 2007 to emphasise the theoretical nature of Lychrel numbers.

## Solution

With Problem 4, we implemented a method to check if a number is a palindrome by reversing the number and if the result is the same as the original number, we know it's a palindrome. As we need the reverse number anyways, we re-implement the principle in Python two functions:

```
def reverse(x):
n = 0
while x > 0:
n = n * 10 + x % 10
x = x // 10
return n
def isPalindrome(n):
return reverse(n) == n
```

We know from the problem statement that the maximum number of iterations for all numbers below 10,000 is \(<50\) to form a Lychrel number. As such the check if a certain number is a Lychrel number is just

```
def isLychrel(n):
for i in range(50):
n+= reverse(n)
if isPalindrome(n):
return False
return True
```

Finally, to check for the full range up to 10,000, we loop over the range and check if the numbers are Lychrel numbers:

```
cnt = 0
for i in range(10000):
if isLychrel(i):
cnt+= 1
print(cnt)
```