**I**f you remember the previous post, I had two interpretations about Le Monde mathematical puzzle #639:

Find all integers with less than 11 digits that are perfect squaresandcan be written as a(a+6), a being an integer.

and:

Find all integers with less than 11 digits that are perfect squaresandcan be written as x concatenated with (x+6), x being an integer.

I got a nice email from Jamie Owen, from Newcastle, Britain, about an R resolution with a clever code, as opposed to mine!

About the second version of the puzzle, Jamie first creates the vector of concatenations:

x = 1:1e5 cats = x * (10^floor(log10(x+6) + 1) +1)+ 6

He then made the function perfect more… perfect:

perfect=function(b){ a=trunc(sqrt(b)) any((a:(a+1))^2 == b) }

(using a function any() I had not seen before, and then got the collection of solutions as

x = 1:1e5 x[sapply(x * (10^floor(log10(x+6) + 1) +1)+ 6,perfect)] [1] 15 38

which runs about 25 times faster than my R solution! (And he further designed a 100 times faster version…)

Jamie also proposed an R code for solving the first version of that puzzle:

max = 1e10 squares = (1:floor(sqrt(max)))^2 # possible answers to a(a+6) a = -1e6:1e6 # which squares have solutions sols = intersect(a*(a + 6), squares) # what are they? f = function(x){ power = floor(floor(log10(x))/2)+1 a = -10^power:10^power sols = c(x,a[a*(a+6) - x == 0]) names(sols) = c("square", "a1", "a2") sols } sapply(sols,f) ## [,1] ## square 16 ## a1 -8 ## a2 2

which returns again 2 as the unique positive solution (equivalent to -8, if considering relative integers). A great lesson in efficient R programming, thanks Jamie!