def euler36(upper_bound=10 ** 6):
"""http://projecteuler.net/index.php?section=problems&id=36
Find the sum of all numbers, less than one million, which are palindromic
in base 10 and base 2."""
p_sum = 0
# note: we can skip even numbers since they're never binary palindromes
for n in range(1, upper_bound, 2):
if is_palindrome(n) and is_palindrome(to_binary(n)):
p_sum += n
return p_sum
def euler125(upper_bound=10**8):
"""http://projecteuler.net/problem=125
The palindromic number 595 is interesting because it can be written as the
sum of consecutive squares: 6^2 + 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2.
There are exactly eleven palindromes below one-thousand that can be written
as consecutive square sums, and the sum of these palindromes is 4164. Note
that 1 = 0^2 + 1^2 has not been included as this problem is concerned with
the squares of positive integers.
Find the sum of all the numbers less than 108 that are both palindromic and
can be written as the sum of consecutive squares.
"""
top_square = int(upper_bound**0.5)
square_sums = [0]*top_square
for i in xrange(1, top_square):
square_sums[i] = square_sums[i-1] + i*i
answers = {}
for i in xrange(1, len(square_sums)):
for j in xrange(i-1):
candidate = square_sums[i] - square_sums[j]
if candidate < upper_bound and is_palindrome(candidate):
answers[candidate] = True
return sum(answers.keys())
def euler4(upper_bound=1000):
"""http://projecteuler.net/problem=4
A palindromic number reads the same both ways. The largest palindrome made
from the product of two 2-digit numbers is 9009 = 91 * 99.
Find the largest palindrome made from the product of two 3-digit numbers.
"""
max_pal = 0
for x in xrange(0, 1000):
for y in xrange(x, 1000):
if x * y > max_pal and is_palindrome(x * y):
max_pal = x * y
return max_pal
def euler55():
"""http://projecteuler.net/problem=55
2520 is the smallest number that can be divided by each of the numbers from
1 to 10 without any remainder.
What is the smallest positive number that is evenly divisible by all of the
numbers from 1 to 20?
"""
lychrel = []
for n in range(10000):
candidate = n
is_lychrel = True
for round in range(50):
candidate += int(str(candidate)[::-1])
if is_palindrome(candidate):
is_lychrel = False
break
if is_lychrel:
lychrel.append(n)
return len(lychrel)
from euler_util import is_palindrome if __name__ == "__main__": count = 0 for n in range(1, 10**4): m = n for i in range(53): m = m + int(str(m)[::-1]) if is_palindrome(str(m)): count += 1 if n == 47: print "OK" break print count
