After the Singapore Maths Olympiad birthday problem that went viral, here is a Vietnamese primary school puzzle that made the frontline in The Guardian. The question is: *Fill the empty slots with all integers from 1 to 9 for the equality to hold*. In other words, find *a,b,c,d,e,f,g,h,i* such that

*a*+13x*b*:*c*+*d*+12x*e*–*f*-11+*g*x*h*:*i*-10=66.

With presumably the operation ordering corresponding to

*a*+(13x*b*:*c)*+*d*+(12x*e)*–*f*-11+(*g*x*h*:*i)*-10=66

although this is not specified in the question. Which amounts to

*a*+(13x*b*:*c)*+*d*+(12x*e)*–*f*+(*g*x*h*:*i)*=87

and implies that *c* divides *b* and *i* divides *g*x*h*. Rather than pursing this analytical quest further, I resorted to R coding, checking by brute force whether or not a given sequence was working.

baoloc=function(ord=sample(1:9)){
if (ord[1]+(13*ord[2]/ord[3])+ord[4]+
12*ord[5]-ord[6]-11+(ord[7]*ord[8]/
ord[9])-10==66) return(ord)}

I then applied this function to all permutations of {1,…,9}* [with the help of the perm(combinat) R function]* and found the 128 distinct solutions. Including some for which b:c is not an integer. (Not of this obviously gives a hint as to how a 8-year old could solve the puzzle.)

As pointed out in a comment below, using the test == on scalars is a bad idea—once realising some fractions may be other than integers—and I should thus replace the equality with an alternative that bypasses divisions,

baoloc=function(ord=sample(1:9)){
return(((ord[1]+ord[4]+12*ord[5]-ord[6]-87)*
ord[3]*ord[9]+13*ord[2]*ord[9]+
ord[3]*ord[7]*ord[8]==0)*ord)}

leading to the overall R code

sol=NULL
perms=as.matrix(data.frame(permutations(9)),ncol=9,byrow=TRUE)
for (t in 1:factorial(9)){
a=baoloc(perms[t,])
if (a[1]>0) sol=rbind(sol,a)}
sol=sol[do.call(order, as.data.frame(sol)),]

and returning the 136 different solutions…