# http://projecteuler.net/index.php?section=problems&id=1
#  look at the first 1000 numbers and add those to the sum,
#  which are multiples of 3 or 5

sum = 0 # initial sum

1000.times{ |n| # iteration index (0-999)
  sum += n if n%3==0 || # add to sum if divisible by
              n%5==0    #  3 or 5 without rest
}

puts sum

# http://projecteuler.net/index.php?section=problems&id=2
#  calculate fibonacci (non-recursive) and sum all even values

n=m = 1 # first two fib. numbers
sum = 0 # initial sum

while m < 4_000_000 # as long as fib. sequence is < 4 000 000
  n,m = m,n+m # calculate next step (sum of the prev. 2 numbers)
  sum += n  if n.even? # add to sum if even [ruby 1.9]
end

puts sum

# http://projecteuler.net/index.php?section=problems&id=3
#  find the largest prime factor of 600 851 475 143

a = 600_851_475_143
f = 2 # smallest possible prime factor

# Divide "evil big number" by lowest prime factors to get the
#  biggest one. Prime factors are those numbers, which do not
#  leave a rest. If a possible number is not a prime factor 
#  candidate anymore, check next one
a%f==0 ? a/=f : f+=1  while a>1

puts f

# http://projecteuler.net/index.php?section=problems&id=4
#  find the largest palindrome made from xxx*xxx

products = # get an array of ALL xxx*xxx
  (100..999).map{ |a|
    (100..999).map{ |b|
      a*b
    }
  }.flatten.select{ |p|      # select those, which are reversed
    p.to_s == p.to_s.reverse # the same as non-reversed
  }.sort.reverse # sort them and reverse result

puts products[0] # put out first (and largest) palindrome

# http://projecteuler.net/index.php?section=problems&id=5
#  find a number that you can divide by 1..20
#  using the method described at the solution sheet at euler

# needed input
till = 20
primes = [2,3,5,7,11,13,17,19]

# get factor for each prime
factors = primes.map{ |f|
  f > (limit ||= till**0.5) ? 1 : # only calcutlate if f < square root of till
  ( Math.log(20)/Math.log(f) ).to_i # calc using log (see euler sheet)
}

# now multiply all primes**factors
number = 1
p = primes.each  # init extern iterators
f = factors.each

loop{ number *= p.next**f.next } # loop breaks when StopIteration is raised

# output
puts number