Final answer:
The values of the digits A, B, and C that solve the equation 24A + 3B6 = C94 are A = 8, B = 4, and C = 5, which is determined through a step-by-step process considering the ones, tens, and hundreds place values respectively.
Step-by-step explanation:
The student's question involves finding the values of the digits A, B, and C in the sum 24A + 3B6 = C94. To solve for these digits, we look at the ones, tens, and hundreds place values and solve the resulting equations.
Step 1: Ones place
Since A + 6 must end with a 4, A must be 8 (as 8 + 6 = 14). We carry over 1 to the tens place.
Step 2: Tens place
Now we have 1 (carried over) + 4 (from 24A) + B must end with 9. This means B must be 4, as 1 + 4 + 4 = 9.
Step 3: Hundreds place
Finally, we have 2 (from 24A) + 3 (from 3B6) adding up to the hundreds place in C94. We don't have any carryover this time, which means C must be the sum of 2 and 3, so C is 5.
Therefore, the digits that replace A, B, and C are 8, 4, and 5 respectively, making the equation 248 + 346 = 594.