Project Euler 179

14 has four divisors: 1, 2, 7 and 14.

14+1, ie. 15, also has four divisors: 1, 3, 5 and 15.

How many integers, n, where 1 < n < 107, has the same number of divisors as n + 1?

That is surprisinly simple. Start by figuring out how many divisors all the numbers have.

``````n <- 10000000
t <- integer(n)

for(s in 1:n){
t[seq(s,n, by=s)] <- t[seq(s,n, by=s)] + 1
}
``````

First I make a vector with the length 107 containing only zeroes.

Element 1 coresponds to n = 1, element 47 to n = 47.

Every single element in that vector has 1 as a divisor.

Every other element has 2 as a divisor.

Every third element has 3 as a divisor, and so on.

So for every element, in the sequence 2 to 107, in steps of 2, I add 1.

The same for every element in the sequence 3 to 107 in steps of 3.

And so on.

That takes a loooong time to do.

Now I have a vector t, where the value in t[n] is the number of divisors in n.

I now need to compare them.

If I have two identical vectors q and r. And add a 0 to the beginning of one of them, and a 0 at the end of the other. I will have two vectors, where element i+1 in the first is the same as element i in the second (or the other way around). So when I compare q[47] with r[47], I actually compare q[47] with q[48]. That is the comparison I need. When these to values are identicial, I have an iteger with the same number of divisors as the following integer.

Apoligies for the not so fluent prose. It’s pretty warm and I’m rather tired.

Anyway:

``````q <- c(0,t)
r <- c(t,0)
``````

I now just need to compare them, and sum the result of that comparison:

``````answer <- sum(q==r)
``````

Lesson learned

Just because a problem has a high number, doesn’t mean that it is particularly difficult.