2 ? 1 ? 6 ? 6 ? 200 ? 50 = 416.56

Let ? be an operator of +,-,*,/,^ that are replaceable (can be used more than once)

How would you solve this math puzzle?

One could potentially work out the math manually, but how would you solve this math problem using R?

1. Set operators as a vector

operators <- c("+","-","*","/","^")

2. Set for loop + monte carlo simulation to fit the equation

for (i in c(1:1000000)) {
  # monte carlo/random sampling of the operator vectors with replacement
  new_op <- sample(operators, 5, replace=T)
  
  # parse the numbers and operators into an expression
  expression <- parse(text = paste0("2",new_op[1],"1",new_op[2],"6",
                                    new_op[3],"6",new_op[4],"200",
                                    new_op[5],"50"))
  
  # evaluate the expression, apparently you cannot just 
  # paste operator and expect R to run it like normal
  sum <- eval(expression) 
  
  # Set condition, if sum is 416.56 then HOT DIGGITY DOG!
  if (sum==416.56) {
    
    # print the answer
    print(paste0("found it, the combinations are: ",expression))
    break
  }  
}

## [1] "found it, the combinations are: 2^1 * 6^6/200 - 50"

Conclusion/Lesson learnt

• Use parse and eval for operators
• simulation is awesome !!! Solved this problem in seconds!