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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+13xb:c+d+12xe–f-11+gxh:i-10=66.

With presumably the operation ordering corresponding to

a+(13xb:c)+d+(12xe)–f-11+(gxh:i)-10=66

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

a+(13xb:c)+d+(12xe)–f+(gxh:i)=87

and implies that c divides b and i divides gxh. 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.)