he product of the two-digit number a4 and the three-digit number 3bc is 7656. What is the average of a, b and c? (a, b, c are digits)

1. ### Break Down the Problem

We have two unknowns in the equation:

• A two-digit number represented as $10a + 4$ (where $a$ is the digit in the tens place).
• A three-digit number represented as $300 + 10b + c$ (where $b$ is the digit in the tens place and $c$ is the digit in the units place).

We need to solve for $a$, $b$, and $c$ given that:

$(10a + 4)(300 + 10b + c) = 7656$

2. ### Relevant Concepts

1. We will expand the product and then solve for feasible values of $a$, $b$, and $c$ that are digits (0-9).
2. We can simplify the problem using divisibility and bounds.

3. ### Analysis and Detail

1. Find possible values of $10a + 4$:
   Since $10a + 4$ is a two-digit number, $a$ can be from 1 to 9. Hence possible values for $10a + 4$ are:

• $14, 24, 34, 44, 54, 64, 74, 84, 94$.

2. Calculate the corresponding $3bc$ for each:
   • We will divide $7656$ by each two-digit number $10a + 4$ to find $300 + 10b + c$: $3bc = \frac{7656}{10a + 4}$

Calculations:

• For $10a + 4 = 14$: $3bc = \frac{7656}{14} = 546 \quad \text{(Not valid, as it is a three-digit number)}$
• For $10a + 4 = 24$: $3bc = \frac{7656}{24} = 319 \quad (3, 1, 9)$
• For $10a + 4 = 34$: $3bc \approx \frac{7656}{34} \approx 225.41 \quad \text{(Not valid)}$
• For $10a + 4 = 44$: $3bc \approx \frac{7656}{44} \approx 174 \quad (1, 7, 4)$
• For $10a + 4 = 54$: $3bc \approx \frac{7656}{54} \approx 142 \quad (1, 4, 2)$

Continuing this way:

From above valid pairs:

• For $10a + 4 = 24$ $\Rightarrow a = 2, b = 1, c = 9$

4. ### Verify and Summarize

• Values are $a = 2$, $b = 1$, and $c = 9$.
• The average of $a$, $b$, and $c$ is calculated as: $\text{Average} = \frac{a + b + c}{3} = \frac{2 + 1 + 9}{3} = \frac{12}{3} = 4$

Final Answer

The average of $a$, $b$, and $c$ is $4$.