This project is intended to document my efforts to learn Node.js, and to maintain my numerical math skills.

Problem 1:

• Tempted to brute force it (either look at each number less that 1000 and check whether it was divisible), but decided to get a bit fancier.
• Essentially this problem is looking for the sum of all numbers less than 1000 divisible by 3 plus the sum of all numbers less than 1000 divisible by 5 minus the sum of all numbers less than 1000 divisible by 15.
• OR ... 3SUM(1..FLOOR(1000/3)) + 5SUM(1..FLOOR(1000/5)) - 15*SUM(1..FLOOR(1000/15))
• SUM(1..N) inclusive = N * ( N + 1 ) / 2... and we're done.

Problem 2:

• A quick browse of wikipedia gives the closed form solution for the sequence: F(n) = ( phi ^ n - (-phi) ^ (-n) ) / sqrt(5)... phi = 1/2 + SQRT(5)/2
• It turns out ever third element is even, so I only need to sum every third element
• Can't see a clever way to solve this, so I'll just run a loop... feels like there should be a better way though

Problem 3:

• Can't think of anything fancy here... I'll hack something together

Problem 4:

• Once again, can't think of anything fancy... I'll at least pass around a function...

Problem 5:

• This one seems easier to solve by hand than with a computer...
• I'll solve for 1..n where n is arbitrary
• Start with the largest number, work downwards.
• Factorize each, and remove the factors from the list, until you get to the smallest element in the list.
• Multiply remaining numbers together, and you're done
• Ouch! This approach is wrong... lets think some more.
• Need to take the prime factors of each number. For each prime factor, need as many of them as the largest requirement
• Multiply them all together (some prime factors more than once)
• After another incorrect submission (!), and a javascript bug is hunted down, this technique seems to work

Problem 6:

• This one seems too easy. I wonder if there is some elegant solution

Problem 7:

• also pretty simple

Problem 8:

• no problem

Problem 9:

• this one looks like fun

Problem 10:

• need an efficient way to get a sequence of primes... 2000000 is largish
• keep a list of all the primes i know of so far, test anything new against those
• wrong answer !?! what have I done wrong?
• whoops, index off by 1

Problem 11:

• Brute force. Dull

Problem 12:

• I like this one. Brute force is NOT going to cut it.
• I can't do a logarithmic search because numDivisors is NOT monotonic in the triangle numbers sequence
• Means I need to do a linear search, which means I need to be VERY efficient (500 divisors will be a large number)
• I already have a reasonably efficient prime factorization function, need to make a function that takes a list of prime factors gives number of divisors

Problem 13:

• Another brute force problem