# Project Euler Q5 :: Smallest multiple

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) # TRUE
```

Easy 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 # 232792560

### CORRECT
```

Wrapping 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
```