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def evaluate_for_primes(primes, count)
	numbers = (primes[count-1]**2+1..primes[count]**2).to_a
	1.upto(count) {|c| numbers.select! {|number| number%primes[c] != 0}}
	primes.concat(numbers)
end

primes, count = [1, 2], 1
until primes.length > 10001 do
	evaluate_for_primes(primes, count)
	count += 1

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# Solution to Project Euler's first problem
# http://projecteuler.net/index.php?section=problems&id=1
# Sum all numbers divisible by 3 or 5 below 1000.

sum = 0
1.upto(999) do |n|
	if (n % 3 == 0 || n % 5 == 0) then 
		sum = sum + n
	end
end

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# Solution to Project Euler's second problem
# http://projecteuler.net/index.php?section=problems&id=2
# Sum all even numbers in the Fibonnacci sequence below 4000000.
i1, i2, sum = 1, 1, 0
until i1 > 4000000
	if i1.even? then sum = sum + i1 end
	i1 = i1 + i2
	if i2.even? then sum = sum + i2 end
	i2 = i2 + i1
end

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# Solution to Project Euler's third problem
# http://projecteuler.net/index.php?section=problems&id=3
# What is the largest prime factor of the number 600851475143?
def generate_primes(ceiling)
	primes, count = [1, 2], 1
	until primes.last > Math.sqrt(ceiling) do
		numbers = (primes[count-1]**2+1..primes[count]**2).to_a
		1.upto(count) {|c| numbers.select! {|number| number%primes[c] != 0}}
		primes.concat(numbers)
		count += 1

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# Solution to Project Euler's fourth problem
# http://projecteuler.net/index.php?section=problems&id=4
# Find the largest palindromic number which is the product of two three-digit numbers.
def generate_and_test
	numbers = []
	999.downto(100) do |x|
		999.downto(100) do |y|
			number = x * y
			if number.to_s == number.to_s.reverse then
				numbers.push number

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# Solution to Project Euler's fifth problem
# http://projecteuler.net/index.php?section=problems&id=5
# Find the smallest number divisible with no remainder by each number below twenty.
def generate_primes(ceiling)
	primes, count = [1, 2], 1
	until primes.last > ceiling do
		numbers = (primes[count-1]**2+1..primes[count]**2).to_a
		1.upto(count) {|c| numbers.select! {|number| number%primes[c] != 0}}
		primes.concat(numbers)
		count += 1

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# Solution to Project Euler's sixth problem
# http://projecteuler.net/index.php?section=problems&id=6
#	Find the difference between the square of the sum of the first hundred natural numbers and the sum of their squares.
def sum_squares(ceiling)
	sum = 0
	(1..ceiling).each {|n| sum = sum +n**2}
	return sum
end
def square_sums(ceiling)
	sum = 0

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# Solution to Project Euler's eighth problem
# http://projecteuler.net/index.php?section=problems&id=8
# Find the largest product of five consecutive digits in a thousand-digit number.
def generate_array(number)
	number = number.to_s
	array = []
	(0..number.length-1).each {|n| array.push number.slice(n).to_i}
	return array
end
def find_products(array)

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# Solution to Project Euler's ninth problem
# http://projecteuler.net/index.php?section=problems&id=9
# Find the product of the Pythagorean triplets whose sum equals 1000.
def find_Pythagorean_triplet(number)
	squares = (1..number).to_a.reverse 
	squares = squares.each_index {|s| squares[s] = squares[s]**2}
	p squares
	squares.each do |c|
		squares.each do |b|
			squares.each do |a|

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# Solution to Project Euler's tenth problem
# http://projecteuler.net/index.php?section=problems&id=10
# Find the sum of all primes below two million.
def generate_primes(ceiling)
	primes, count = [1, 2], 1
	until primes.last > ceiling do
		numbers = (primes[count-1]**2+1..primes[count]**2).to_a
		1.upto(count) {|c| numbers.select! {|number| number%primes[c] != 0}}
		primes.concat(numbers)
		count += 1