A 2-digit number is such that the sum of the number and the number obtained by reversing the order of the digits of the number is 55. Further, the difference of the given number and the number obtained by reversing the order of the digits of the number is 45. What is the product of the digits?
0
Representing the 2-Digit Number
Let's represent the unknown 2-digit number. We can denote the digit in the tens place as '\(a\)' and the digit in the units place as '\(b\)'.
- The value of the 2-digit number can be written as \(10a + b\).
- When the order of the digits is reversed, the new number formed has '\(b\)' in the tens place and '\(a\)' in the units place. Its value is \(10b + a\).
For example, if the number is 72, then \(a=7\) and \(b=2\). The reversed number is 27 (\(10 \times 2 + 7\)).
Setting Up Equations for Sum and Difference
The problem gives us two conditions:
- Sum Condition: The sum of the original number and the reversed number is 55.
Mathematically, this is: \((10a + b) + (10b + a) = 55\) Simplifying this equation: \(11a + 11b = 55\) We can divide both sides by 11: \(a + b = 5 \quad \textbf{(Equation 1)}\) - Difference Condition: The difference between the original number and the reversed number is 45.
Mathematically, this is: \((10a + b) - (10b + a) = 45\) Simplifying this equation: \(9a - 9b = 45\) We can divide both sides by 9: \(a - b = 5 \quad \textbf{(Equation 2)}\)
Solving for the Digits
Now we have a system of two linear equations with two variables:
- \(a + b = 5\)
- \(a - b = 5\)
We can solve this system. One way is to add the two equations together:
Adding Equation 1 and Equation 2:
\( (a + b) + (a - b) = 5 + 5 \) \( 2a = 10 \)Divide by 2 to find '\(a\)':
\( a = \frac{10}{2} \) \( a = 5 \)Now, substitute the value of '\(a\)' (which is 5) back into Equation 1 to find '\(b\)':
\( 5 + b = 5 \) \( b = 5 - 5 \) \( b = 0 \)So, the digits of the 2-digit number are \(a=5\) and \(b=0\). The number is 50.
Calculating the Product of Digits
The question asks for the product of the digits.
The digits are \(a=5\) and \(b=0\).
The product is \(a \times b\).
\( \text{Product} = 5 \times 0 \) \( \text{Product} = 0 \)Verification
Let's check if the number 50 satisfies the conditions:
- The number is 50. The reversed number is 05 (or 5).
- Sum: \(50 + 5 = 55\). This matches the first condition.
- Difference: \(50 - 5 = 45\). This matches the second condition.
Both conditions are met.
Final Answer
The product of the digits of the 2-digit number is 0.
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The third proportional to 9 and 15 is:
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