Solving the Repeating Digit Number Problem

This problem asks us to evaluate an expression involving numbers with repeating digits. The key is to find a way to represent these numbers algebraically so that we can perform calculations easily. We are given three numbers, x, y, and z, defined by sequences of repeating digits:

• x consists of 20 digits of '1'.
• y consists of 10 digits of '3'.
• z consists of 10 digits of '2'.

We need to calculate the value of the expression $(x – y^2)/z$.

Representing Repeating Digit Numbers

A number consisting of $n$ digits of '1' can be written as the sum of powers of 10: $10^{n–1} + 10^{n–2} + ... + 10^1 + 10^0$. This is a geometric series. The sum of this series is $\frac{10^n - 1}{10-1} = \frac{10^n - 1}{9}$.

A number consisting of $n$ digits of 'k' is simply $k$ times a number consisting of $n$ digits of '1'. So, it can be represented as $k \times \frac{10^n - 1}{9}$.

Expressing x, y, and z Algebraically

• x = 111...1 (20 digits): Using the formula, $x = \frac{10^{20} - 1}{9}$.
• y = 333...3 (10 digits): Using the formula with k=3 and n=10, $y = 3 \times \frac{10^{10} - 1}{9} = \frac{10^{10} - 1}{3}$.
• z = 222...2 (10 digits): Using the formula with k=2 and n=10, $z = 2 \times \frac{10^{10} - 1}{9}$.

Calculating the Expression (x – y²)/z

First, let's calculate the numerator: $x – y^2$.

Substitute the expressions for x and y:

$x – y^2 = \frac{10^{20} - 1}{9} – \left(\frac{10^{10} - 1}{3}\right)^2$

$x – y^2 = \frac{10^{20} - 1}{9} – \frac{(10^{10} - 1)^2}{3^2}$

$x – y^2 = \frac{10^{20} - 1}{9} – \frac{(10^{10} - 1)^2}{9}$

Combine the terms over a common denominator:

$x – y^2 = \frac{(10^{20} - 1) – (10^{10} - 1)^2}{9}$

Expand the term $(10^{10} - 1)^2$ using the identity $(a-b)^2 = a^2 - 2ab + b^2$:

$(10^{10} - 1)^2 = (10^{10})^2 - 2 \times 10^{10} \times 1 + 1^2 = 10^{20} - 2 \times 10^{10} + 1$

Substitute this back into the numerator expression:

$x – y^2 = \frac{(10^{20} - 1) – (10^{20} - 2 \times 10^{10} + 1)}{9}$

Remove the parentheses in the numerator (remembering to change the signs):

$x – y^2 = \frac{10^{20} - 1 – 10^{20} + 2 \times 10^{10} – 1}{9}$

Combine like terms:

$x – y^2 = \frac{(10^{20} - 10^{20}) + 2 \times 10^{10} + (–1 – 1)}{9}$

$x – y^2 = \frac{0 + 2 \times 10^{10} – 2}{9} = \frac{2 \times 10^{10} – 2}{9} = \frac{2(10^{10} – 1)}{9}$

Now, let's calculate the full expression $(x – y^2)/z$. Substitute the calculated value of the numerator and the expression for z:

$\frac{x – y^2}{z} = \frac{\frac{2(10^{10} - 1)}{9}}{2 \times \frac{10^{10} - 1}{9}}$

To divide by a fraction, we multiply by its reciprocal:

$\frac{x – y^2}{z} = \frac{2(10^{10} - 1)}{9} \times \frac{9}{2(10^{10} - 1)}$

Assuming $10^{10} - 1 \neq 0$ (which is true), we can cancel out the common terms in the numerator and denominator:

$\frac{x – y^2}{z} = 1$

Thus, the value of $(x – y^2)/z$ is 1.

Summary of Calculation Steps

1. Express x, y, and z using the formula for repeating digits: $k \times \frac{10^n - 1}{9}$.
2. Substitute these expressions into $(x – y^2)/z$.
3. Simplify the numerator $x – y^2$ by expanding $y^2$ and combining terms.
4. Divide the simplified numerator by the expression for z.
5. Cancel common factors to get the final result.

Revision Table: Repeating Digit Number Calculation

Reviewing the key components of the problem and solution helps reinforce the concepts.

Additional Information on Number Patterns

Problems involving repeating digits are common in competitive exams and number theory. Understanding how to represent these numbers in a compact form using powers of 10 is a valuable skill. Here are some related points:

• Numbers like 1, 11, 111, etc., form a sequence where the n-th term is $\frac{10^n - 1}{9}$. These are sometimes called Repunits (for digit 1).
• Numbers like 33, 333, etc., can be seen as 3 times the corresponding repunit. Similarly, 22, 222, etc., are 2 times the repunit.
• This algebraic representation simplifies operations like addition, subtraction, multiplication, and division of such large repeating digit numbers.
• Recognizing that $10^{20} - 1$ can be related to $10^{10} - 1$ using the difference of squares or by recognizing the structure $10^{20} - 1 = (10^{10})^2 - 1^2 = (10^{10} - 1)(10^{10} + 1)$ is also helpful in simplifying such problems, although the direct expansion of $(10^{10} - 1)^2$ worked directly here.