abdollar

SIXTYTWO = ('0'..'9').to_a + ('a'..'z').to_a + ('A'..'Z').to_a

def to_s_62(i)
  return '0' if i == 0
  s = ''
  while i > 0
    s << SIXTYTWO[i.modulo(62)]
    i /= 62
  end
  s.reverse

abdollar

#!/usr/bin/env ruby

# A simply utility to show character counts for each line of input and
# highlight lines longer than 80 characters.
#
# Written as an example for http://jstorimer.com/2011/12/12/writing-ruby-scripts-that-respect-pipelines.html
#
# Examples:
#
# $ hilong Gemfile

abdollar

# ruby getimage.rb -e production -c config/getimage.rb

abdollar

fib=->(n){_=5**0.5;(((1+_)**n-(1-_)**n)/(2**n*_)).to_i}

abdollar

# Square root calculation using Newton-Raphson
# from Practical Programming (1968)

a = 256

x = (1 + a) / 2.0

loop do
  ox = x
  x = (x + a.to_f / x) / 2.0

abdollar

require 'prime'
Prime.take(10001).last

abdollar

# Convert ssh-rsa key to pem
ssh-keygen -f infile.pub -e -m PKCS8 > outfile.pem

# Encrypt a file using public key pem
openssl rsautl -encrypt -inkey public.pem -pubin -in file.txt -out file.ssl

# Decrypt using private key
openssl rsautl -decrypt -inkey private.pem -in file.ssl -out decrypted.txt

abdollar

#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

ruby-1.9.2-p180 :004 > (1..10).to_a.reduce(:+)**2 - (1..10).to_a.map { |x| x**2 }.reduce(:+)
 => 2640 
ruby-1.9.2-p180 :005 > (1..100).to_a.reduce(:+)**2 - (1..100).to_a.map { |x| x**2 }.reduce(:+)

abdollar

(400..2432902008176640000).step(20).each do |i|
   val = (1..20).to_a.select{|x| i%x==0}&(1..20).to_a
   if val.count == 20
     puts i
     exit
   end
end

Use the product of the LCM's

abdollar

def palindrome?(str); str.reverse == str; end

max = 0
900.upto(999) { |a|
  a.upto(999) { |b|
    max = [max, a * b].max if palindrome?((a * b).to_s)
  }
}
max