Here's a systematic way of solving this type of problem. It exhausts all the possibilities given the specified constraints. At the end I offer "exotic solutions".
There are 4 numbers to be combined, so split them into two pairs. We can have (5,5) and (6,12): (5,6) and (5,12). We now take each pair and list all permissible binary operations between them which result in a positive number:
PAIR 1
(5,5): 5×5=25, 5➗5=1, 5+5=10, 5-5=0, 55;
(6,12): 6×12=72, 6➗12=½, 12➗6=2, 6+12=18, 12-6=6, 126, 612.
PAIR 2
(5,12): 5×12=60, 5➗12=5/12, 12➗5=2⅖, 12+5=17, 12-5=7, 125, 512;
(5,6): 5×6=30, 5➗6=⅚, 6➗5=1⅕, 6+5=11, 6-5=1, 56, 65.
We then take the results from the first row of Pair 1 and try to combine each one with each of the results from the second row of Pair 1. For example, 2 from the second row subtracted from the 25 from row 1=23, which is close to 24, but isn't equal to it. There are 35 possible combinations of the two rows and 4 permissible operations (multiply, divide, add, subtract) to apply to each combination to get the number 24. The same exercise is then repeated for Pair 2. For example, 2⅖×11=132/5=26⅖, larger than 24, where 12➗5=2⅖ and 6+5=11. There are 49 possible combinations.
There seems to be no solution so far. The nearest is 5×5-6➗12=24½.
The next step is to consider grouping three out of the four numbers using two binary operations and combine the result with the remaining number.
These are the 3 triads (groups of 3): (5,5,6), (5,5,12), (5,6,12). As an example, take (5,6,12). Split this into ((5○6)○12) and (5○(6○12)) where ○ represents any one of the 4 arithmetic operators. We already have all possible results for 5○6 and 6○12 in the pair rows, so we simply take the third member of the triad and combine it with the pair row results (duplicate pair row results are omitted):
TRIAD 1 (5,5,6)
((5○5)○6)={25, 1, 10, 0, 55}○6=25×6=150, 25/6, 25+6=31, 25-6=19; 6, ⅙, 7, 5; 60, 5/3, 16, 4; 0; 330, 6/55, 61, 49;
(5○(5,6))=5○{30, ⅚, 6/5, 11, 1, 56, 65}=150, ⅙, 6, 35, 25; 25/6, ⅙, 35/6; 6/25, 31/5, 19/5; 55, 5/11, 11/5, 16; 5, ⅕, 4; 280, 5/56, 56/5, 61, 51; 325, 1/13, 13, 70, 60.
TRIAD 2 (5,5,12)
((5○5)○12)={25, 1, 10, 0, 55}○12=300, 25/12, 12/25, 37, 13; 12, 1/12, 13, 11; 120, ⅚, 6/5, 22, 2; 0; 660, 55/12, 12/55, 67, 43;
(5○(5,12))=5○{60, 5/12, 12/5, 17, 7, 125, 512}=300, 1/12, 12, 65, 55; 25/12, 65/12, 55/12; 12/25, 37/5, 13/5; 85, 5/17, 17/5, 22; 35, 5/7, 7/5, 2; 625, 1/25, 25, 130, 120; 2560, 5/512, 512/5, 517, 507.
TRIAD 3 (5,6,12)
((5○6)○12)={30, ⅚, 6/5, 11, 1, 56, 65}○12=360, 5/2, ⅖, 42, 18; 10, 5/72, 72/5, 77/6, 67/6; 66/5, 54/5; 132, 11/12, 12/11, 23, 1; 1/12, 12, 13, 11; 672, 14/3, 3/14, 68, 44; 780, 65/12, 12/65, 77, 53;
(5○(6,12))=5○{72, ½, 2, 18, 6, 126, 612}=360, 5/72, 72/5, 77, 67; 5/2, 10, 1/10, 11/2, 9/2; ⅖, 7, 3; 90, 5/18, 18/5, 23, 13; 30, ⅚, 6/5, 11, 1; 630, 5/126, 126/5, 131, 121; 3060, 5/612, 612/5, 617, 607.
We now need to apply the combined results of each triad with its specific "remainder" from the set of 4 numbers. For example, we take the whole set of results from both rows of Triad 1 and apply the 4 arithmetic operators to those results and the remainder 12. We need only look for results that give us 24. In the case of Triad 1 and remainder 12, we need 2, 288, ½, 12 or 36 in either row. For Triad 2, the remainder is 6, so we would need 4, 144, ¼, 18 or 30 in either of the two rows to make 24. For Triad 3, the remainder is 5, so we'd need 24/5, 120, 5/24, 19 or 29.
(126-5)/5=24⅕ which is closest to 24. There do not appear to be any exact solutions.
EXOTIC SOLUTIONS (bending the rules)
12√(5+5-6)=24
(5+5+6-12)!=24
