Problem Analysis:

We are tasked to solve the problem where two two-digit numbers, $AB$ and $CD$, add up to a three-digit number $1CE$. Here:

• $AB = 10A + B$
• $CD = 10C + D$
• $1CE = 100 + 10C + E$

The digits $A, B, C, D, E$ are distinct.

We must determine the value of $A$, the tens digit of $AB$.

Steps:

1. Formulate the equation for the sum: $AB + CD = 1CE$ Substituting the expanded forms: $(10A + B) + (10C + D) = 100 + 10C + E$ Simplify: $10A + B + 10C + D = 100 + 10C + E$ Cancel $10C$ from both sides: $10A + B + D = 100 + E$ Rearrange: $10A = 100 + E - B - D$

2. Key Observations:

   • Since $AB$ and $CD$ are two-digit numbers, $A \geq 1$ and $C \geq 1$.
   • The result $1CE$ must start with $1$, so $100 \leq AB + CD < 200$.

3. Range for $A$: From the equation: $10A = 100 + E - B - D$ The right-hand side must be divisible by 10. Since $E, B,$ and $D$ are distinct digits, possible values for $A$ must be checked.

Trial-and-Error with Logical Elimination:

Let us consider $AB + CD = 1CE$ while ensuring the constraints on distinct digits.

• $A = 9$: $10(9) + B + D = 100 + E$ This simplifies to: $B + D = E$ A valid $B, D,$ and $E$ configuration is possible.

• $A = 8$ or $7$: Testing these values does not yield consistent results for $AB + CD = 1CE$ within the distinct digit constraint.

Conclusion:

The only feasible value for $A$ is 9, so the correct answer is:

(a) 9

Would you like me to explain any specific step in more detail or test alternative scenarios? Let me know!

Follow-Up Questions:

1. How do we ensure that all digits $A, B, C, D, E$ are distinct?
2. Why must $AB + CD$ fall between 100 and 200?
3. Could there be other solutions with the same $A$ value?
4. How would the solution change if the digits were not distinct?
5. What is the significance of divisibility by 10 in simplifying the equation?

Tip:

When solving digit-based problems, always consider the constraints of the number system (e.g., distinct digits, valid ranges) to narrow down possibilities systematically.