  Calculation Challenge Game: Some Statistics
 All of the statistics on this page are specific to the Calculation Challenge Game when the Goal integer = 24 and the factor range is 1 to 13.    SaneSchool
 Excellent work done quickly.    As the table above shows, there are more valid solutions if one of the factors is 1, 2, 3, 4, 6, 8, or 12 and fewer valid solutions if one of the factors is 5, 7, 9, 10, 11, or 13, which seems pretty reasonable. Note:  In the table above, there are a total of 1276248 + 217089120 + 983112 = 219348480 possible equations with 4 factors, in which each of the 4 factors can be 1 to 13.  We can verify this total by considering how many possible equations we would expect:  There are 13 possible numbers for each of the 4 factors and 7680 ways to combine them using 4 operations (+, -, x, /) and 2 sets of parentheses, which gives 134 x 7680 = 219348480 expected equations.  (We’ve calculated the number of possible equations, not the number of unique equations!) Unsolvable factor sets
 There are 459 unique unsolvable factor sets (listed here).  These are sets of factors in which no combination of adding, subtracting, multiplying, or dividing the 4 factors results in 24.  The table below shows the number of unsolvable sets that contain at least one of the factors from 1 to 13:
 As the table above shows, over a third of the unsolvable sets contain at least one factor equal to 13.  Additionally, over a quarter of the unsolvable sets contain a factor of 9, 10, 7, or 11, with the factor of 1 very nearly in that category.  At the other extreme, only 13% of the unsolvable sets contain a 2 or a 3. We state elsewhere that the largest number of valid solutions occurs when the factors are all 12s.  In general though, having all the same factor is more likely to be an unsolvable set of factors.  Of the possible 13 factor sets with all the same factors, only 5 of them have valid solutions (those in which the factors are all 3, 4, 5, 6, or 12); the remaining 8 factor sets are unsolvable (those in which the factors are all 1, 2, 7, 8, 9, 10, 11, or 13). So, how many unique combinations of 4 numbers are there in which each number can be 1 to 13? There are 5 cases to consider:                  Case 1:  None of the 4 numbers are repeated                                   13 x 12 x 11 x 10 = 17160                  Case 2:  1 number is repeated and the other 2 numbers are different                                   (13 x 1) x 12 x 11 = 1716                  Case 3:  2 numbers are repeated                                   (13 x 1) x (12 x 1) = 156                  Case 4:  1 number is repeated 3 times and the 4th number is different                                   (13 x 1 x 1) x 12 = 156                  Case 5:  1 number is repeated 4 times                                   (13 x 1 x 1 x 1) = 13 Adding up these 5 cases, we calculate that there are 17160 + 1716 + 156 + 156 + 13 = 19201 unique factor sets of 4 numbers. We can use the above results to calculate that 459 / 19201 = 2.39% of the factor sets are unsolvable, which means that 97.61% are solvable.  In other words, don’t be in a hurry to give up on finding a solution—nearly all of the factor sets are solvable!
 If one of the factors is this... there are this many valid solutions that equal 24, this many valid equations that don’t equal 24, and this many equations that are invalid due to divide-by-zero errors. 1 117912 16670856 84192 2 121824 16671468 79668 3 114456 16680552 77952 4 114120 16681668 77172 5 81660 16715532 75768 6 111612 16685424 75924 7 74964 16723788 74208 8 104838 16693758 74364 9 81870 16717506 73584 10 80682 16718850 73428 11 78564 16722060 72336 12 117288 16682556 73116 13 76458 16725102 71400 Total 1276248 217089120 983112
 This many unsolvable sets of factors contain at last one of this factor. 60 2 60 3 75 4 75 12 83 6 97 5 99 8 113 1 123 9 132 10 138 7 148 11 162 13