Basildon Coder

Onwards to…

Problem 6

The sum of the squares of the first ten natural numbers is,

1² + 2² + … + 10² = 385

The square of the sum of the first ten natural numbers is,

(1 + 2 + … + 10)² = 55² = 3025

Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 385 = 2640.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

Bit of a disappointment, problem 6; it’s too easy. It’s rated as the third-easiest, i.e. easier than problems 3, 4, and 5 which I’ve already covered. In fact, for my money it’s easier than problem 2 as well. Ah well, the difficulty ramps up soon enough, trust me.  Here’s the very simple python solution:

sum_sq = sum([ x*x for x in xrange(1, 101)])
sq_sum = sum(xrange(1, 101)) ** 2

print sq_sum - sum_sq

As you can see, it’s pretty intuitive. You sum the squares, square the sum, and calculate the difference. The answer is basically in the description, you just have to scale up a little.

There’s not much else to say about this one. Even if I abandon the functional approach and write a straightforward imperative solution it’s still very straightforward. In (deliberately non-idiomatic, so don’t whine at me) ruby:

sum_of_squares = 0
sum = 0

1.upto 100 do |x|
    sum_of_squares += x * x
    sum += x
end

p (sum * sum) - sum_of_squares

This entry was posted on Thursday, August 14th, 2008 at 12:00 am and is filed under Coding, Mathematics, Project Euler, Software Engineering. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.