Happy New Year

Yesterday’s post from Programming Praxis asks us to build a very special sort of expression. Using the numbers 10, 9, 8, 7, 6, 5, 4, 3, 2, and 1 in that order along with the operators of multiplication, division, addition, subtraction, and concatenation, find all of the ways that we can write an expression totaling 2013. Here’s one valid solution:

109 - 8 * 7 + 654 * 3 - 2 / 1 = 2013


Project Euler 5

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? – PROJECT EULER #5


Project Euler 4

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

Find the largest palindrome made from the product of two 3-digit numbers. – PROJECT EULER #4


Project Euler 2

Each new term in the Fibonacci sequence is generated by adding the previous two terms.

By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …

By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms. – Project Euler #2


Project Euler 1

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000. – Project Euler #1


The Sum Of The First Billion Primes

This problem from Programming Praxis came about in the comments to my last post and intrigued me. So today, we are trying to sum the first one billion primes. Summing the first hundred, thousand, even million primes isn’t actually that bad. But it takes a bit more effort when you scale it up to a billion. And why’s that?


Pandigital Sums

Yesterday’s new post from Programming Praxis asked us to find all triples (a, b, a+b) such that a and b are three digits and a+b is four and concatenating the numbers results in a pandigital number (one with all 10 digits). After that, find the smallest individual number in any of these triples.


Pythagorean Triples

When Programming Praxis mentioned that the newest challenge sounded like a Project Euler problem, they were’t wrong. Basically, the idea is to count the number of Pythagorean Triples with perimeters (sum of the three numbers) under a given value. The necessary code to brute force the problem is really straight forward, but then they asked for the count up to one million. With the brute force O(n^2) algorithm (and a relatively high constant), that’s not really feasible. So that’s when we have to get a bit more creative.