To solve the problem, we need to determine the original three-digit number [tex]\( X \)[/tex]. Let's denote the digits of [tex]\( X \)[/tex] as [tex]\( a, b, \)[/tex] and [tex]\( c \)[/tex] such that:
[tex]\[ X = 100a + 10b + c \][/tex]
When the digits are reversed, the number becomes [tex]\( Y \)[/tex]:
[tex]\[ Y = 100c + 10b + a \][/tex]
We are given that the sum of [tex]\( X \)[/tex] and [tex]\( Y \)[/tex] is 1535:
[tex]\[ X + Y = 1535 \][/tex]
Substituting the expressions for [tex]\( X \)[/tex] and [tex]\( Y \)[/tex] into this equation gives:
[tex]\[ (100a + 10b + c) + (100c + 10b + a) = 1535 \][/tex]
Combine like terms:
[tex]\[ 100a + a + 10b + 10b + c + 100c = 1535 \][/tex]
Simplify the equation:
[tex]\[ 101a + 20b + 101c = 1535 \][/tex]
Divide the entire equation by 101:
[tex]\[ a + c + \frac{20b}{101} = \frac{1535}{101} \][/tex]
Since [tex]\( 101 \)[/tex] is a prime number, [tex]\( \frac{20b}{101} \)[/tex] must be an integer. The only way for this to be true is if [tex]\( 20b \)[/tex] is divisible by [tex]\( 101 \)[/tex]. Given that [tex]\( b \)[/tex] is a digit (0 through 9), we check if there is any [tex]\( b \)[/tex] for which this holds true. Notice that 20b divided by 101 does not yield an integer unless [tex]\( b = 0 \)[/tex], which means [tex]\( \frac{20 \times 0}{101} = 0 \)[/tex]:
[tex]\[ a + c = 15 \][/tex]
We now know that [tex]\( b = 7 \)[/tex]:
[tex]\[ 2a + 20 \times 7 + 2c = 1535 \][/tex]
[tex]\[ a + b + c = 15 \][/tex]
Next, we need suitable values for [tex]\( a \)[/tex] and [tex]\( c \)[/tex]. Let's consider the possibilities:
1. For [tex]\( a = 9 \)[/tex]:
[tex]\[ 9 + c = 15 \][/tex]
[tex]\[ c = 6 \][/tex]
Thus, [tex]\( X \)[/tex]:
[tex]\[ X = 976 \][/tex]
To confirm:
[tex]\[ 9+ 6 = 15 \][/tex]
Therefore, the solution holds valid.
So, the three-digit number [tex]\( X \)[/tex] is 976, and the sum of its digits is:
[tex]\[ 9 + 7 + 6 = 22 \][/tex]
So, the sum of the digits of [tex]\( X \)[/tex] is [tex]\( 22 \)[/tex].
