If the 9-digit number 5x82y6588 is exactly divisible by `99(x+ylt10)`, then what is the value of `y-x`?

Options

A. 4
B. `-4`
C. 1
D. `-1`

Answer

To solve the problem, we need to determine the values of \( x \) and \( y \) in the 9-digit number \( 5x82y6588 \) such that it is divisible by \( 99 \). Since \( 99 = 9 \times 11 \), we need to check for divisibility by both \( 9 \) and \( 11 \).

### Step 1: Check for divisibility by 9
The rule for divisibility by \( 9 \) states that the sum of the digits must be divisible by \( 9 \).

The digits of the number are \( 5, x, 8, 2, y, 6, 5, 8, 8 \).

Calculating the sum of these digits:
\[ 5 + x + 8 + 2 + y + 6 + 5 + 8 + 8 = 42 + x + y \]

For this sum to be divisible by \( 9 \):
\[ 42 + x + y \equiv 0 \mod 9 \]

Calculating \( 42 \mod 9 \):
\[ 42 \div 9 = 4 \quad \text{(remainder } 6\text{)} \]
So, \( 42 \equiv 6 \mod 9 \).

Thus, we need:
\[ 6 + x + y \equiv 0 \mod 9 \implies x + y \equiv 3 \mod 9 \]

### Step 2: Check for divisibility by 11
The rule for divisibility by \( 11 \) states that the difference between the sum of the digits in the odd positions and the sum of the digits in the even positions must be divisible by \( 11 \).

The digits in odd positions are \( 5, 8, y, 5, 8 \) (1st, 3rd, 5th, 7th, 9th):
\[ 5 + 8 + y + 5 + 8 = 26 + y \]

The digits in even positions are \( x, 2, 6, 8 \) (2nd, 4th, 6th, 8th):
\[ x + 2 + 6 + 8 = x + 16 \]

Now, we need:
\[ (26 + y) - (x + 16) \equiv 0 \mod 11 \]
This simplifies to:
\[ 10 + y - x \equiv 0 \mod 11 \implies y - x \equiv 1 \mod 11 \]

### Step 3: Solve the equations
We have two equations:
1. \( x + y \equiv 3 \mod 9 \)
2. \( y - x \equiv 1 \mod 11 \)

From the second equation, we can express \( y \) in terms of \( x \):
\[ y = x + 1 \]

Substituting \( y \) into the first equation:
\[ x + (x + 1) \equiv 3 \mod 9 \]
\[ 2x + 1 \equiv 3 \mod 9 \]
\[ 2x \equiv 2 \mod 9 \implies x \equiv 1 \mod 9 \]

The possible values for \( x \) (since \( x \) must be a single digit) are \( 1 \) or \( 10 \). Since \( x \) must be less than \( 10 \), we have:
\[ x = 1 \]

Substituting \( x = 1 \) back to find \( y \):
\[ y = 1 + 1 = 2 \]

### Step 4: Calculate \( y - x \)
Now we can find \( y - x \):
\[ y - x = 2 - 1 = 1 \]

Thus, the final answer is:
\[ \boxed{1} \]