Problem: if w is the set of all the integers between 49 and 114 , inclusive , that are either multiples of 3 or multiples of 2 or multiples of both , then w contains how many numbers ?
Rationale: "official solution : number of multiples of 3 step 1 . subtract the extreme multiples of 3 within the range ( the greatest is 114 , the smallest is 51 ) : 114 - 51 = 63 step 2 . divide by 3 : 63 / 3 = 21 step 3 . add 1 : 21 + 1 = 22 . so there are 22 multiples of 3 within the range : examples are 51 , 54 , 57 , 60 , etc . number of multiples of 2 step 1 . subtract the extreme multiples of 2 within the range ( the greatest is 114 , the smallest is 50 ) : 114 - 50 = 64 step 2 . divide by 2 : 64 / 2 = 32 step 3 . add 1 : 32 + 1 = 33 . so there are 33 multiples of 2 within the range : examples are 50 , 52 , 54 , 56 , 58 , 60 etc . add the 22 multiples of 3 and the 33 multiples of 2 : 22 + 33 = 55 . however , by adding the multiples of 2 and the multiples of 3 , we are effectively counting several numbers twice : for example , 54 and 60 are parts of both the lists above . so we ca n ' t just take 22 + 33 = 55 . find the number of multiples of 6 ( which are the double counted , as 6 is divisible by both 2 and 3 ) , and subtract it from 55 : step 1 . subtract the extreme multiples of 6 within the range ( the greatest is 72 , the smallest is 54 ) : 114 - 54 = 60 step 2 . divide by 6 : 60 / 6 = 10 step 3 . add 1 : 10 + 1 = 11 . so there are 11 multiples of 6 within the range : we counted 11 numbers twice . subtract the 11 multiples of 6 from the sum of the multiples of 2 and 3 : = 22 + 33 - 11 = 55 - 11 = 44 therefore , the final number of multiples of 2 , 3 or 6 is 44 . hence , this is the correct answer . ( b )"
annotated_formula: subtract(add(floor(divide(subtract(114, 49), 3)), divide(subtract(114, 49), 2)), floor(divide(subtract(114, 49), multiply(2, 3))))