Project Euler problem 6 is one of the easiest that I have encountered so far. Also, Problem 7 is quite easy to solve once you have a function for prime numbers (something we did in Problem 3). Therefore, in this post I give 2 problems for the price of 1! :)
Let's have a look at the question for Problem 6:
The sum of the squares of the first ten natural numbers is,
1^2 + 2^2 + ... + 10^2 = 385
The square of the sum of the first ten natural numbers is,
(1 + 2 + ... + 10)^2 = 55^2 = 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.
There are a few known tricks to calculate the sum of the squares; however, given the limited size of the problem at hand, brute-force is perfectly fine: it shows once more the elegance of a declarative solution.
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
What is the 10001st prime number?
As I mentioned in the introduction, this one is fairly easy once you know how to tell whether a given number is prime or not. We did that exercise in Problem 3.
There's a catch though: this time we are counting the prime numbers.
In the test we did in Problem 3
there was a small error that in that context was irrelevant: the test returned
also for 0 and 1, which strictly speaking are not prime numbers (2 is considered to be the first prime number).
For this reason, I have modified the code of
a bit so that the first prime will be 2, the second 3, and so on as appropriate.
Ok... break time is over. Problem 8 looks a bit more challenging. See you when I have got a nice solution for that.