Project Euler in Ruby

p1.rb

puts (1...1000).inject(0) {|sum,i| 
	sum = (sum + i) if i%5 == 0 or i%3 == 0
	sum
 }

p10.rb

# Find the sum of all the primes below two million.

max = 2_000_000

# # 3rd approach: optimized sieve
@@primes = Array.new(max,true)

# cross out even nrs
4.step(max,2) do |n|
	@@primes[n] = false
end
# now we start at 3
sum = 2

3.step(max,2) do |n|
	if @@primes[n]
		sum += n 
		# sieve n mults!
		# You only need to start crossing out multiples at p^2 , because any smaller multiple of p has a prime divisor less than p and has already been crossed out as a multiple of that
		(n*n).step(max, 2*n) do |s| # step 2*n because even products are already out
			@@primes[s] = false
		end
	end
end

puts sum

# 2nd approach: sieve. faster for big limits
# @@non_primes = []
# sum = 0
# 
# (2..max).each do |n|
# 	unless @@non_primes[n]
# 		sum += n 
# 		# sieve n mults!
# 		(n*2).step(max,n) do |s|
# 			@@non_primes[s] = true
# 		end
# 	end
# end
# 
# puts sum

# 1st approach: trial division. not too bad with the sqrt(n) limit
# @@primes = []
# 
# def prime?(n)
# 	sqrt = Math.sqrt(n).to_i
# 	# if n isnt prime, it MUST divide by one of the lower primes
# 	@@primes.each do |p|
# 		break if p > sqrt # improvement: we may stop at sqrt(n)
# 		return false if n%p == 0
# 	end
# 	# if it doesnt, it's a new prime
# 	@@primes << n
# 	return true
# end
# 
# # we're skiping the 2 so we can skip even numbers in the cycle
# sum = 2
# @@primes << 2
# 
# 3.step(max,2) do |n|
# 	sum += n if prime?(n)
# end
# 
# puts sum

p2.rb

pp = 0
p = 1
c = 1
sum = 0
loop do
	# current value
	c = p + pp
	# too big?
	break if c > 4000000
	# sum when due
	sum += c if c%2 == 0
	# shift values
	pp = p
	p = c
end

puts sum

p3.rb

n = 600851475143

(2..n).each do |f|
	break if f>n # done! - n is updated inside the cycle, this will happen
	if n%f == 0
		puts f
		n = n/f
	end
end

p4.rb

def pal?(n)
	s = n.to_s
	(0..s.size/2).each do |c|
		return false if s[c] != s[s.size-c-1]
	end
	return true
end

largest = 0

999.downto(100) do |i|
	999.downto(100) do |j|
		p = i*j
		break if p < largest
		if pal?(p)
			largest = p
		end
	end
end

puts largest

p5.rb

puts (1..20).inject(1) {|t,n| t.lcm n}

p6.rb

# euler - 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.

# brute force. is there another way?
sum = 0
sq_sum = 0
1.upto(100) do |n|
	sum += n
	sq_sum += n**2
end
puts sum**2 - sq_sum 

# yes there is. from the forums:
#    The sum of squares is n*(n +1)*(2*n +1) /12
#    The square of sum of n numbers is  ( n*(n+1)/2 ) * ( n *(n +1)/2 )
#    If we take the diffrence and solve it algebraically
#    We get the diffrence to be  -
#        ( 3 * n^4 + 2 * n^3 -3 * n^2 - 2 * n )/12
#        so answer is the value of above expression
#    To reduce number of multiplication operations  store value of  n^2 .
#

p7.rb

@@primes = []

def prime?(n)
	# if n isnt prime, it MUST divide by one of the lower primes
	@@primes.each do |p|
		return false if n%p == 0
	end
	# if it doesnt, it's a new prime
	@@primes << n
	return true
end

nth = 10001
count = 0
n = 2

loop do 
	count += 1 if prime?(n)
	break if count == nth
	n += 1
end

puts n

p8.rb

n1k = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450"

max = 0

(0..n1k.size-5).each do |index|
	prod = n1k[index,5].chars.inject(1) {|s,e| s*e.to_i}
	max = prod if prod > max
end

puts max

p9.rb

	1.upto(1000) do |a|
		1.upto(1000) do |b|
			csq = a**2 + b**2
			c = Math.sqrt(csq)
			next if c%1 != 0 # not an int
			puts "#{a} #{b} #{c} #{a*b*c}" if a+b+c == 1000
		end
	end