Numbers with equal mean and median

Let x, y & z be the three positive numbers chosen such x>y>z.

It is obvious therefore that y shall be the median.

Making y also as the mean of the other two leads us to the equation

(x+z)/2=y or x+z=2y.

The sum of x and z shall then be even, happening when x & z are both even or both odd.

Thus, the number of ways shall be:

both even: 7C2 (the numbers 2, 4, 6, 8, 10, 12, & 14 taken two at a time); and

both odd: 8C2 (the numbers 1, 3, 5, 7, 9, 11, 13, & 15 taken two at a time).

=> 7C2 + 8C2 = 21 + 28 = 49.

The total number of ways of choosing three distinct positive integers is 15C3 = 455.

Therefore, the probability shall be a/b = 49/455 = 7/65. a+b=72 :)

Suppose that the three integers are $x, y, z$ , and further suppose that $0 < x < y < z \le 15$ . Clearly, there are $\binom{15}{3} = \frac{15\cdot14\cdot13}{3\cdot2} = 5\cdot7\cdot13$ ways to choose $x, y, z$ . We want the mean to equal the median. That is, we want $\frac{x+y+z}{3} = 3y$ to be satisfied. This is equivalent to $x + y + z = 3y$ , or $x + z = 2y$ .

If $y = 1$ , then there are no $(x, z)$ that meet all the conditions. If $y = 2$ , then the possible $(x, z)$ is $(1, 3)$ . Note that for each $y$ up to $y = 8$ , there will be $y-1$ pairs of $(x, z)$ that satisfy the conditions: $(1, 2y-1)$ , $(2, 2y - 2)$ , $\cdots$ , $(y-1, y+1)$ . For $y > 8$ , these cases are equivalent to the cases symmetric about $y = 8$ . That is, $y = 9$ is the same as $y = 7$ , $y = 10$ is the same as $y = 6$ , and so forth up to $y = 15$ is the same as $y = 1$ (try it and see for yourself if this isn't clear). Thus, the total number of ways for the conditions to be satisfied is $2(1+2+3+4+5+6) + 7 = 2(21) + 7 = 49$ .

The probability that the conditions are satisfied is $\frac{49}{5 \cdot 7 \cdot 13} = \frac{7}{65}$ , and the desired answer is $7 + 65 = \boxed{72}$ .