Final answer:
The equation ab * bc = cac is invalid for all digits a, b, and c from 1-9, as the multiplication cannot generate the same digit c in the ones place and it cannot hold the same tens digit a after adding a carried digit b.
Step-by-step explanation:
The question asks for the values of digits a, b, and c such that when a and b form a two-digit number ab and b and c form bc, their product equals the three-digit number cac. To find the solution, we need to consider the multiplication of two-digit numbers and recognize the place values involved.
Starting with the equation ab * bc = cac, we know that b * c must end with the digit c, as that is the ones digit in the product cac. This implies that c must be 1, since any other digit would not recreate itself when multiplied by another digit. With c being 1, the equation simplifies to ab * b1 = ca1.
Next, the tens digit in the product ca1 must come from a * 1 + b, where b is carried over from the multiplication of b with the ones place of b1. The only way for a to remain the same after adding b is for b to be 0, which is not permitted as all digits are significant and must be from 1-9. Therefore, this multiplication is impossible given the criteria, and there must be a misunderstanding or typo in the question as written.