Multiple Mystery

Relevant wiki: Postage Stamp Problem / Chicken McNugget Theorem

Since $a| lcm(a, b)$ & $a| lcm(c, a)$ , then $a| lcm(b, c)$ , for all these multiples are equal.

Then we can use Dr. Warm's formula for the Frobenius number given:

$g(a, b, c) = lcm(a, c) + lcm(a, b) - a - b -c = 29$

Now from the constraint, suppose $x = lcm(a, b) = lcm(b, c) = lcm(c, a)$ . Then $a + b + c = x + 1$ .

Hence, $29 = 2x - (x + 1) = x -1$ .

Thus, $x = 30 = lcm(a, b) = lcm(b, c) = lcm(c, a)$ .

Now since $30 = 2\cdot 3\cdot 5$ , there are only $3$ prime factors for multiple combination.

The number $30$ can not be applicable for $c$ , for the sum of $a+b =1$ is not possible as they are positive integers.

By trying all possible factors, it will be obvious that the triple $(a, b, c) = (6, 10, 15)$ as it is the only combination, where $a + b + c - 1 = 6+10+15-1 =30$ .

As a result, $lcm(a-1, b-1, c-1) = lcm(5, 9, 14) = \boxed{630}$ .