Explanation. Standard caveat: don’t look here if you are trying to do these yourself.
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?
I’m getting the feeling that brute-force is going to be quite the useful tool for these questions. Thankfully R can churn through numbers really fast.
So, we’re after a number divisible by 1, 2, 3, ..., 10. Let’s vectorise that and check the stated answer
all(2520 %% 1:10 == 0)## [1] TRUEEasy enough. The solution value must be divisible by 20, so we can just test multiples of 20 for the above property
i <- 20
y <- FALSE
while(!y) {
i <- i + 20
y <- all(i %% 1:20 == 0)
}
i## [1] 232792560### CORRECTWrapping a system.time() call around that assures us that this is still done in under a minute, as per the guidelines
user system elapsed
26.150 0.000 26.192
devtools::session_info()
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