Project Euler Part 2

August 13, 2021

Welcome back!

Today we are continuing our journey through the Euler Project.

4th problem:

A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.

Find the largest palindrome made from the product of two 3-digit numbers.

So let’s break this into smaller problems:

a) A function to detect if a number is palindrome.

b) A function that gets a list of possible products of the range of two numbers.

(defn get-list-of-products [n1 n2]
  (for [n1 (range n1)
        n2 (range n2)]
    (* n1 n2))

c) A function that gives us a number with a certain amount of digits.(Optional, I could just hardcode the 999). Here I’m just thinking something like (10^(ammount-of-digits))-1.

So I wrote unit test for each and implemented them.

Our main function will do the following in order:

1. Create two numbers with three digits (999). (function c)
2. Get a list of possible products of two number (function b)
3. Remove duplicates of the list (Optional, this is for performance I don’t want to iterate on the list on unnecessary values.)
4. Filter the list to only those who are a palindrome.
5. Sort from the lowest number to the highest.
6. Take the last value.

Tada 🎉!

While I was implementing function c I learned that there isn’t a exponential function in the cloure core library, but instead I had to install a math library [org.clojure/math.numeric-tower "0.0.4"].

My biggest take away was lazy sequence and taking advantage of it in function b where instead of looping using a stateful index I just leverage range n that returns a list from 0 up to n. On top of this lazy sequence I use nested for loop to make a combination of all the possible values of two list!

Clojure is so elegant!

5th Problem

2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

So from what I got from this problem is that 2520 is the lowest number that can be divided by [1 2 3 4 5 6 7 8 9 10] and the remainder is 0 for every division.

So if we want to get the lowest number we probably can start a 10 and go on to 11 after that lets call this variable n.

So we loop over n increasing it each iteration. If at any time we find a n that has a remainder of 0 when dividing against every? [1 2 3 4 5 6 7 8 9 10] then we return the number!

Straight forward! Only problem, my algorithm took almost 70 sec to find the lowest posible number who is divisible by [1-20]! Read over the Euler Project rules, and it seems its fine, for now.

Onwards!!

6th Problem

Hmmm you know what we do here:

• A function that returns the sum of the squares of the first n natural numbers.
• A function that returns the squares of sums of the first n natural numbers.
• A function that uses these two functions and gets the difference.

I love how Euler gives you like test samples which are great for the test for each individual function above. Its like they also believe in TDD!

Straight forward!

Onwards!

Written by Edgardo Carreras.

• ← Project Euler Part I
• Project Euler Part 3 →