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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
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