Given a number x with n digits that ends in the digit 2. The digit 2 that is in the last place (the unity digit - right) - is moved to the left (to the left end of the number) and you get 2x. FindThe original X

The problem can be stated mathematically as follows:

You are given a number \( x \), which has \( n \) digits and ends in the digit 2. After moving the last digit (2) to the leftmost position, the new number formed is \( 2x \).

Let’s break it down step-by-step:

### Step 1: Represent \( x \)
Assume \( x \) has \( n \) digits and ends with the digit 2. So, we can write:
\[
x = 10y + 2
\]
where \( y \) is the number formed by the first \( n-1 \) digits of \( x \).

### Step 2: After moving the last digit (2) to the front
When we move the digit 2 from the last position to the first position, the new number becomes \( 2 \times 10^{n-1} + y \). According to the problem, this number is equal to \( 2x \). Therefore, we can set up the following equation:
\[
2 \times 10^{n-1} + y = 2x
\]

### Step 3: Substituting \( x \)
Substitute \( x = 10y + 2 \) into the equation:
\[
2 \times 10^{n-1} + y = 2(10y + 2)
\]
Simplify the right-hand side:
\[
2 \times 10^{n-1} + y = 20y + 4
\]

### Step 4: Solve for \( y \)
Rearranging terms:
\[
2 \times 10^{n-1} - 4 = 19y
\]
Thus:
\[
y = \frac{2 \times 10^{n-1} - 4}{19}
\]

### Step 5: Solve for \( x \)
Now that we have \( y \), substitute it back into the expression for \( x \):
\[
x = 10y + 2
\]

So, \( x \) can be found using the formula:
\[
x = 10 \times \frac{2 \times 10^{n-1} - 4}{19} + 2
\]

### Example: \( n = 3 \)
Let’s work through an example where \( n = 3 \) (a 3-digit number). The equation becomes:
\[
y = \frac{2 \times 10^{3-1} - 4}{19} = \frac{2 \times 100 - 4}{19} = \frac{196}{19} = 10
\]
Then, \( x = 10 \times 10 + 2 = 102 \).

If we move the last digit of 102 (which is 2) to the front, we get 210, and indeed \( 2 \times 102 = 210 \).

### Conclusion
The formula to find the original number \( x \) is:
\[
x = 10 \times \frac{2 \times 10^{n-1} - 4}{19} + 2
\]

This works for any \( n \)-digit number that satisfies the problem condition.

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Would you like further details on this solution, or perhaps explore more examples? Here are five related questions:

1. How does the solution change if the last digit is different from 2?
2. Can you prove that this method works for all values of \( n \)?
3. What happens if \( n = 1 \)? Does the method still hold?
4. What is the relationship between the original number and the new number after moving the last digit?
5. Can this method be generalized to other transformations of digits?

**Tip:** When manipulating digits in mathematical problems, break the number into parts (e.g., powers of 10) to make equations easier to handle.

Given a number x with n digits that ends in the digit 2. The digit 2 that is in the last place (the unity digit - right) - is moved to the left (to the left end of the number) and you get 2x. Find the original X.

Solve a mathematical problem where a number x with n digits ends with the digit 2, and after moving the 2 to the front, the result is 2x. Learn the algebraic steps, including using powers of 10 and rearranging equations to find the original number x. Suitable for Grades 9-12.

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