Multiply the natural numbers

Divide these numbers into three groups:(2,6,10,14,18,22), (4,8,12,16,20,24), third group consists of remaining numbers. So group of three numbers that isn't divisible by 4 can be selected from third group or choosing two from third group and one from first group. The number of groups that are not divisible by 4 is: $\binom{12}{3}+6\binom{12}{2}=616$ Total number of groups is: $\binom{24}{3}=2024$ $\frac{a}{b}=\frac{1408}{2024}=\frac{16}{23}$ $a+b=\boxed{39}$

There are $24\cdot 23\cdot 22$ triples $(a,b,c) \in \{1, \dots, 24\}^2$ without repetition. The product $abc$ is not a multiple of four if

• $a, b,c$ are all odd

• or one of them is even, the others are odd.

The number of triples that satisfy the first condition is $n_1 = 12\cdot 11\cdot 10.$ The number of triples for which $a$ is even (but no multiple of 4) and $b, c$ are odd is $n_2 = 6\cdot 12\cdot 11.$ Likewise for triples with only $b$ even, or only $c$ even.

Thus there are $n_1 + 3n_2 = (10 + 18)\cdot 11\cdot 12 = 28\cdot 11\cdot 12$ triples such that $abc$ is not a multiple of four. The desired probability is $1 - \frac{28\cdot 11\cdot 12}{24\cdot 23\cdot 22} = 1- \frac{28}{4\cdot 23} = 1 - \frac 7{23} = \frac{16}{23}.$ Thus the answer is $16 + 23 = \boxed{39}$ .

We have 12 odd numbers and 12 even numbers, out of which we have to consider 3 numbers in either of the following ways:

a) 3 odds;

b) 2 odds & 1 even;

c) 1 odd & 2 evens;

d) 3 evens;

Case (a) never satisfies our criteria.

Case (c) can be done in $^{12}C_{1}$ x $^{12}C_{2}$ ways.

Case (d) can be done in $^{12}C_{3}$ ways.

Case (b) can satisfy our criteria if the only even number is a multiple of 4, which can be done in $^{12}C_{2}$ x $^{6}C_{1}$ ways.

Total number of ways of selecting 3 numbers out of 24 is $^{24}C_{3}$ .

So the desired probability is =

$\frac{(12*66) + (220) + (66*6)}{2024} = \frac{16}{23}$ which means a+b = 39