Understanding Two-Digit Numbers and Their Digits

A two-digit number can be represented algebraically using its digits. Let the tens digit be \(a\) and the units digit be \(b\). Since it's a two-digit number, \(a\) must be a non-zero integer from 1 to 9, and \(b\) must be an integer from 0 to 9.

The value of the number is given by \(10 \times (\text{tens digit}) + 1 \times (\text{units digit})\). So, the number is \(10a + b\).

The sum of the digits is \(a + b\).

When the digits are interchanged, the new number has \(b\) as the tens digit and \(a\) as the units digit. The value of this new number is \(10b + a\).

Setting up the Equations

The problem gives us two conditions based on the original number, the interchanged number, and the sum of the digits. Let's translate these conditions into equations:

1. The original two digit number is \(k\) times the sum of its digits.
   \(10a + b = k(a + b)\)   (Equation 1)
2. The number formed by interchanging the digits is \(m\) times the sum of the digits.
   \(10b + a = m(a + b)\)    (Equation 2)

We are asked to find the value of \(m\) in terms of \(k\).

Solving for m in Terms of k

We have two equations involving \(a\), \(b\), \(k\), and \(m\). A common technique when you have expressions like \((a+b)\) in both equations is to add or subtract the equations.

Let's add Equation 1 and Equation 2:

\((10a + b) + (10b + a) = k(a + b) + m(a + b)\)

Combine like terms on the left side:

\(10a + a + b + 10b = k(a + b) + m(a + b)\)

\(11a + 11b = k(a + b) + m(a + b)\)

Factor out 11 from the left side and \((a+b)\) from the right side:

\(11(a + b) = (k + m)(a + b)\)

Since \(a\) is the tens digit of a two-digit number (\(a \neq 0\)) and \(b\) is the units digit, their sum \((a+b)\) cannot be zero. Therefore, we can divide both sides of the equation by \((a + b)\):

\(\frac{11(a + b)}{a + b} = \frac{(k + m)(a + b)}{a + b}\)

\(11 = k + m\)

Now, we can solve for \(m\) by subtracting \(k\) from both sides:

\(m = 11 - k\)

Conclusion

The value of \(m\) is \(11 - k\). This relationship holds true for any two-digit number satisfying the given conditions, as long as the sum of the digits is not zero.

Revision Table: Two-Digit Number Concepts

Additional Information: Number Properties

This problem demonstrates a useful property of two-digit numbers and their digit sums. The sum of a two-digit number and the number formed by reversing its digits is always 11 times the sum of the digits.

\((10a + b) + (10b + a) = 11a + 11b = 11(a + b)\)

Similarly, the difference between a two-digit number and the number formed by reversing its digits is always 9 times the difference of the digits (tens digit minus units digit).

\((10a + b) - (10b + a) = 10a + b - 10b - a = 9a - 9b = 9(a - b)\)

These properties are often useful in solving problems involving two-digit numbers and their digits.