def is_lychrel(n):
count = 1
n += int(str(n)[::-1])
while not is_palindrome(n) and count < 50:
n += int(str(n)[::-1])
count += 1
return not is_palindrome(n)
def largest_palindrome_product(minimum=100, maximum=999):
"""
exhaustively finds the largest palindromic product in a range
minimum, maximum (int): defines range, inclusive
returns (int): solution
"""
stub = Stub()
#stub.start()
candidate_solution = None
palindrome_products = []
candidate_factors = [None, None]
for i in range(maximum, minimum-1, -1):
for j in range(i, minimum-1, -1):
product = i*j
if is_palindrome(product):
palindrome_products.append(product)
if product > candidate_solution:
candidate_solution = product
candidate_factors = [i, j]
stub.msg(candidate_factors, "factors")
return candidate_solution
def is_lychrel(n):
i = 0
while i < 50:
n += int(''.join(list(reversed(digits(n)))))
if is_palindrome(digits(n)):
return False
i += 1
return True
def main():
count = 0
for i in range(1, 10000):
for _ in range(50):
i = step(i)
if euler.is_palindrome(i):
break
else:
count += 1
return count
def main():
maximum = 0
for i in range(10**DIGITS - 1, 10**(DIGITS - 1) - 1, -1):
for j in range(10**DIGITS - 1, i - 1, -1):
prod = i * j
if prod > maximum:
if euler.is_palindrome(prod):
maximum = prod
else:
break
return maximum
def main():
s = set()
for b in range(1, 10000):
temp = b * (b + 1) * (2 * b + 1)
for a in range(b - 2, -1, -1):
result = (temp - a * (a + 1) * (2 * a + 1)) // 6
if result > LIMIT:
break
if euler.is_palindrome(result):
s.add(result)
return sum(s)
def compute():
ways = {}
for i in range(1, int(limit ** 0.5)):
for j in range(1, int(limit ** 0.33)):
temp = i ** 2 + j ** 3
if is_palindrome(temp):
if temp in ways:
ways[temp] += 1
else:
ways[temp] = 1
return sum([w for w in ways if ways[w] == 4][:5])
def main():
total = 0
for n in range(1000000):
if euler.is_palindrome(str(n)) and euler.is_palindrome("{0:b}".format(n)):
total += n
print(total)
# -*- coding:utf-8 -*-
"""Project Euler problem 36"""
from euler import is_palindrome
sm = 0
for i in range(1000000):
if is_palindrome(str(i)) and is_palindrome(str(bin(i))[2:]):
sm += i
print("Answer: " + str(sm))
from euler import is_palindrome
print max(i * j for i in xrange(999, 900, -1) for j in xrange(999, i, -1)
if is_palindrome(i * j))
# coding=utf-8 ''' Problem 36 31 January 2003 The decimal number, 585 = 10010010012 (binary), is palindromic in both bases. Find the sum of all numbers, less than one million, which are palindromic in base 10 and base 2. (Please note that the palindromic number, in either base, may not include leading zeros.) ''' import euler print sum(n for n in range(1, 10**6+1) if (euler.is_palindrome(n) and euler.is_palindrome(bin(n)[2:])))
'''
from euler import is_palindrome
multiples = set()
# Creating the set of all 3-digit multiples.
for factor1 in range(100, 1000):
# For the second loop, we do not need to search and number lower
# than the number we're currently on in the first loop. We have already
# multiplied those two in a previous loop.
# E.g. when factor1 = 102, we do not need to multiply it by factor2 = 101
# because when factor1 = 101, there was an interation when factor2 = 102.
# 102 * 101 = 101 * 102
for factor2 in range(factor1, 1000):
multiples.add(factor1 * factor2)
smallest_multiple = 100 * 100
largest_multiple = 999 * 999
palindromes = set()
# Creating the set of all palindromes.
for number in range(smallest_multiple, largest_multiple):
if is_palindrome(number) == True:
palindromes.add(number)
both = multiples.intersection(palindromes)
answer = max(both)
print(f"the largest palindromic product of two 3-digit numbers is {answer}.")
from euler import is_palindrome
total = 0
for n in range(1000000):
if is_palindrome(str(n)) and is_palindrome(str(bin(n))[2:]):
total += n
print(total)
def palin_gen():
for j in xrange(999, 900, -1):
for k in xrange(999, 900, -1):
n = j * k
if euler.is_palindrome(n):
yield n
from euler import is_palindrome print max(i * j for i in xrange(999, 900, -1) for j in xrange(999, i, -1) if is_palindrome(i * j))
def compute():
return sum([n for n in range(10**6) if is_palindrome(n) and is_palindrome(str(bin(n))[2:])])
import sys
sys.path.append("..")
from euler import is_palindrome
upper = 1000000
double_palindromes = []
for i in range(upper):
digits = list(str(i))
if is_palindrome(digits):
base2 = list(format(i, 'b'))
if is_palindrome(base2):
double_palindromes.append(i)
result = sum(double_palindromes)
print(double_palindromes)
print(result)
def test_is_palindrome(self):
"""Basic palindrome test"""
self.assertTrue(euler.is_palindrome(1001))
import sys
sys.path.append("..")
from euler import is_palindrome
largest = 9009
for i in range(100, 1000):
for j in range(i, 1000):
product = i * j
if product > largest and is_palindrome(product):
largest = product
print(largest)
# -*- coding:utf-8 -*-
"""Project Euler problem 55"""
from euler import is_palindrome
c = 0
for i in range(10000):
p = i
for j in range(50):
p = p + int(str(p)[::-1])
if is_palindrome(p):
break
else:
c += 1
print("Answer: " + str(c))
# -*- coding:utf-8 -*-
"""Project Euler problem 55"""
from euler import is_palindrome
c = 0
for i in range(10000):
p = i
for j in range(50):
p = p + int(str(p)[::-1])
if is_palindrome(p):
break
else:
c += 1
print("Answer: " + str(c))
#!/usr/bin/env python
#Find the largest palindrome made from the product of two 3-digit numbers.
import euler
product = 0
palindrome = 0
for n in range(999, 100, -1):
for x in range(100, 999, 1):
product = n * x
if euler.is_palindrome(str(product)) == True:
if product > palindrome:
palindrome = product
print(palindrome)
def compute():
return max([i * j for i in range(100, 1000) for j in range(100, 1000) if is_palindrome(i * j)])
#!/usr/bin/env python #Find the largest palindrome made from the product of two 3-digit numbers. import euler product = 0 palindrome = 0 for n in range(999,100,-1): for x in range(100,999,1): product = n * x if euler.is_palindrome(str(product)) == True: if product > palindrome: palindrome = product print(palindrome)
