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Project Euler Solutions

Project Euler 125 Solution

Project Euler 125 Solution

Project Euler 125: Finding square sums that are palindromic


Problem Description

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.

Analysis

Build sequences of consecutive squares and record those that are palindromic. Once the sum has exceeded our limit, L, we can break from the current sequence starting number and start with the next. We use a set to eliminate duplicate sum-of-squares.

Project Euler 125 Solution

Runs < 0.350 seconds in Python 2.7.
from Euler import is_palindromic

L = 10**8
sqrt_limit = int(L ** 0.5)    
pal = set()

for i in range(1, sqrt_limit-1):
    sos = i*i
    for j in xrange(i+1, sqrt_limit):
        sos += j*j
        if sos >= L: break
        if is_palindromic(sos): pal.add(sos)

print "Answer to PE125 =", sum(pal), len(pal)
Use this link to get the Project Euler 125 Solution Python 2.7 source.

Answer

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Project Euler 125 Solution last updated July 12, 2014

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