We are given a 3-digit number, represented as XYZ. This means the number can be written algebraically using its digits:
We are also given its reverse, ZYX. This number can be written as:
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.
Let's calculate D using the algebraic representations:
Now, we simplify the expression by combining like terms:
We can factor out 99 from the expression:
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:
Simplifying this gives:
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.
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