Euler Problem 5 relates to the divisibility of numbers.

## Euler Problem 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?

## Solution

The solution will also be divisible by the number 1 to 10 so we can start at 2520 and increment by 2520. The loop checks whether the number is divisible by the numbers 1 to 20.

# Start as high as possible i <- 2520 # Check consecutive numbers for divisibility by 1:20 while (sum(i%%(1:20)) != 0) { i <- i + 2520 # Increase by smallest number divisible by 1:10 } answer <- i

Here’s an implementation using the Euclidean algorithm.

Very nice work David. I did not think to use an analytical approach. I have never used the Reduce function before – thanks for the lesson.