Multiplying 192 with 1, 2 and 3:

192 x 1 = 192

192 x 2 = 384

192 x 3 = 576

Concatenating the products, we get 192384576. That number is 1-9 pandigital (as in all digits 1 to 9 is represented in the number exactly one time).

The same can be achieved by multiplying 9 with 1, 2, 3, 4 and 5. That gives us the pandigital concatenation 918273645.

We should now find the largest 1-9 pandigital number, formed as a concatenation in the described way, by multiplying an integer with 1, 2,…n, where n > 1.

Testing with 918273645. Nope, that is not the solution. That would be too easy 🙂

OK. Lets call the integer we’re multiplying, m.

And as we’re using R, lets call the numbers we are multiplying it with, for n, as in the vector 1:n.

The concatenated product will have to have 9 as its first digit. The first digits comes from multiplying m with 1. Therefore, the first digit in m must be 9.

We need to end up with a total of 9 digits in the concatenated product. If m has two digits, multiplying with 1 will give us two digits in the result. Multiplying with 2 will give us 3 digits in the result (any two digit number larger than 50 will, multiplied by 2, give a three digit result. And we know that m should begin with 9).

Multiplying with 3 will give another 3 digits in the result, bringing us to at total of 8. Multiplying with 4 brings the total number of digits in the concatenated result to 11. There can be only 9.

Chosing a three digit m gives similar problems.

But a four digit m, with 9 as the first digit, gives us four digits multiplying by 1 and five when we multiply by 2. Giving us a total of 9 digits in the result.

What we know so far:

• The integer we have to multiply has four digits, and begins with 9.
• We will have to multiply it by 1 and 2. Or 1:2.

That should be possible to brute-force.

Multiply all four digit integers where the first digit is 9 by 1:2.

Collapse the results to a single string.

Keep all the mumbers that are pandigital.

Take the maximum of it.

I don’t need to check all the numbers. I know that the first four digits must be larger than 9182, since this is the first four digits in the example. The highest number is not 9999, it is 9876.

t <- 9183:9876

Using the purrr library:

library(purrr)
answer <- t %>%
  keep(function(x) length(unique(setdiff(unlist(strsplit(paste(x*1:2,collapse=""),split="")),0)))==9)

Ok, lets stop there. What am I doing?

I pass all the possible four digit integers to keep(). keep() applies a function to each of the integers, and keeps the ones where the function evaluates to true.

The function in question is an anonymous function of x:

length(unique(setdiff(unlist(strsplit(paste(x*1:2,collapse=“”),split=“”)),0)))==9

x, the integer from t, is multiplied wiht 1:2. That results in a vector where the first element is x multiplyed by 1, and the second x multiplied by 2.

That is then put into the paste() function, with the parameter collapse=“”. The result is that the vector from just before, is concatenated to just one string.

That string is splitted by strsplit into individual characters. The result is a list. Which is bloody annoying to work with.
So I put everything into unlist(), that returns a vector.

Setdiff(x,0) removes any zeroes from that vector.

Having removed the zeroes, unique() returns all the unique values in the vector. If there is 9, the number is 1 to 9 pandigital.

And if it is, the integer is kept by keep.

It doesn’t really give me the result. It returns the four digit integers beginning with 9, that, multiplied by 1:2 and concatenated gives a pandigital result.

But the largest of them is the integer I should multiply by 1:2. So lets expand the code a bit:

answer <- t %>%
  keep(function(x) length(unique(setdiff(unlist(strsplit(paste(x*1:2,collapse=""),split="")),0)))==9) %>%
  max() %>%
  (function(x) x*1:2) %>%
  (function(x) paste(x, collapse="")) %>%
  as.numeric()

max() gives me the largest of the four digit integers. The next function mulitplies it by 1:2. After that, the result is collapsed. And finally converted to numeric by as.numeric(). Not really necessary, but I’m doing it anyway.

Lessons learned

When using pipes to do stuff, anonymous functions are useful. But should be put into parantheses.