#!/usr/bin/python
# -*- coding: utf-8 -*-
#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.
#Answer:
#906609
from time import time; t=time()
from mathplus import palindrome
ret = 0
for i in range(999, 99, -1):
for j in range(i, 99, -1):
x = i * j
if x <= ret: break
if palindrome(x):
ret = x
print(ret)#, time()-t
#!/usr/bin/python
# -*- coding: utf-8 -*-
#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.
#Answer:
#906609
from time import time
t = time()
from mathplus import palindrome
ret = 0
for i in range(999, 99, -1):
for j in range(i, 99, -1):
x = i * j
if x <= ret: break
if palindrome(x):
ret = x
print(ret) #, time()-t
#!/usr/bin/python
# -*- coding: utf-8 -*-
#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.)
#Answer:
#872187
from time import time; t = time()
from mathplus import palindrome
M = 1000000
s = 0
for i in range(1, M, 2):
if not palindrome(i): continue
m = bin(i)[2:]
if int(m[::-1], 2) == i:
#print i
s += i
print(s)#, time()-t
#!/usr/bin/python
# -*- coding: utf-8 -*-
#The palindromic number 595 is interesting because it can be written as the sum of consecutive squares: 62 + 72 + 82 + 92 + 102 + 112 + 122.
#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 = 02 + 12 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.
#Answer:
#2906969179
from time import time; t=time()
from mathplus import isqrt, palindrome
M = 10**8
s = isqrt(M/2)
pool = [0]*(s+1)
for i in range(s):
pool[i+1] = pool[i] + i*i
ss = set()
for u in range(s-1):
for v in range(u+2, s+1):
x = pool[v] - pool[u]
if x >= M: break
if palindrome(x): ss.add(x)
print(sum(ss)-1)#, time()-t
