Understanding the Problem: 3-Digit Number Difference

We are given a 3-digit number, represented as XYZ. This means the number can be written algebraically using its digits:

• \(XYZ = 100X + 10Y + Z\)

We are also given its reverse, ZYX. This number can be written as:

• \(ZYX = 100Z + 10Y + X\)

Here, X, Y, and Z represent the digits of the number. X is the hundreds digit, Y is the tens digit, and Z is the units digit. For XYZ to be a 3-digit number, X must be non-zero (\(X \in \{1, 2, ..., 9\}\)). Z can be any digit (\(Z \in \{0, 1, ..., 9\}\)). Similarly, for ZYX to be considered, Z is the hundreds digit and X is the units digit.

The problem defines D as the difference between XYZ and ZYX.

Calculating the Difference (D)

Let's calculate D using the algebraic representations:

• \(D = XYZ - ZYX\)
• \(D = (100X + 10Y + Z) - (100Z + 10Y + X)\)

Now, we simplify the expression by combining like terms:

• \(D = 100X + 10Y + Z - 100Z - 10Y - X\)
• Group the terms involving X, Y, and Z:
• \(D = (100X - X) + (10Y - 10Y) + (Z - 100Z)\)
• \(D = 99X + 0 - 99Z\)
• \(D = 99X - 99Z\)

We can factor out 99 from the expression:

• \(D = 99(X - Z)\)

Finding the Remainder when D is Divided by 99

The question asks for the remainder when D is divided by 99. We have found that \(D = 99(X - Z)\).

Let's perform the division:

• \(\frac{D}{99} = \frac{99(X - Z)}{99}\)

Simplifying this gives:

• \(\frac{D}{99} = X - Z\)

Since X and Z are digits, their difference (\(X - Z\)) is an integer. For example, if X=7 and Z=3, \(X-Z = 4\). If X=2 and Z=8, \(X-Z = -6\). The division \(\frac{99(X - Z)}{99}\) results in the integer \((X - Z)\) with no fractional part.

In mathematical terms, when a number is expressed as \(99 \times k\) (where k is an integer, in our case \(k = X-Z\)), dividing it by 99 always results in the integer k, and the remainder is 0.

Therefore, the remainder when D is divided by 99 is 0.