The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?
Step-by-Step Solution
Key Concept: Represent the two‑digit number using its tens and units digits, form two linear equations from the given conditions (sum of the number and its reversal, and the difference of the digits), and solve the simultaneous equations to obtain the digits. The absolute difference condition gives two possible ordered pairs, leading to two numbers.
1. Let the digits be
- Tens digit = $x$ ( $1 \le x \le 9$ )
- Units digit = $y$ ( $0 \le y \le 9$ )
2. Write the number and its reversal
- Original number $N = 10x + y$
- Reversed number $R = 10y + x$
3. Use the first condition (sum = 66)
\[
N + R = (10x + y) + (10y + x) = 11(x + y) = 66
\]
Hence
\[
x + y = \frac{66}{11} = 6 \quad\text{(Equation 1)}
\]
4. Use the second condition (digits differ by 2)
The absolute difference is 2, so we have two possibilities:
- $x - y = 2$ \(\text{Case A}\)
- $y - x = 2$ \(\text{Case B}\)
5. Solve Case A
\[
\begin{cases}
x + y = 6 \\
x - y = 2
\end{cases}
\]
Adding the equations: $2x = 8 \Rightarrow x = 4$.
Substituting in $x + y = 6$: $4 + y = 6 \Rightarrow y = 2$.
Hence the number is $10x + y = 42$.
6. Solve Case B
\[
\begin{cases}
x + y = 6 \\
y - x = 2
\end{cases}
\]
Adding the equations: $2y = 8 \Rightarrow y = 4$.
Substituting in $x + y = 6$: $x + 4 = 6 \Rightarrow x = 2$.
Hence the number is $10x + y = 24$.
7. Verification
- $42 + 24 = 66$ (satisfies the sum condition).
- Digits differ by $|4-2| = 2$ (satisfies the difference condition).
Both numbers satisfy all the given conditions.
8. Conclusion
The two‑digit numbers are $42$ and $24$. Therefore, the total number of such numbers is 2.
Correct Answer: The numbers are 42 and 24. Hence, there are 2 such numbers.