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The sum of the digits of a three-digit number is eight. The middle digit is thrice the sum of the other two digits. The difference between the number and the number obtained by reversing the order of the digits is zero. What is the number?

A. 161
B. 242
C. 192
D. 125

Sagot :

To solve this problem, we need to determine the three-digit number based on the given conditions. Let's denote the number as [tex]\( \overline{abc} \)[/tex], where [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] are the digits of the number.

### Step-by-Step Solution

1. Sum of the digits:
The sum of the digits [tex]\( a, b, \)[/tex] and [tex]\( c \)[/tex] is given as 8.
[tex]\[ a + b + c = 8 \][/tex]

2. Middle digit relation:
The middle digit [tex]\( b \)[/tex] is three times the sum of the other two digits [tex]\( a \)[/tex] and [tex]\( c \)[/tex].
[tex]\[ b = 3(a + c) \][/tex]

3. Equality of the number and its reverse:
The number is the same as the number obtained by reversing the order of the digits. Hence, [tex]\( \overline{abc} = \overline{cba} \)[/tex].
This means:
[tex]\[ 100a + 10b + c = 100c + 10b + a \][/tex]
Simplifying this equation:
[tex]\[ 99a = 99c \implies a = c \][/tex]

4. Substitute [tex]\( a = c \)[/tex] into the previous equations:
Since [tex]\( a = c \)[/tex], we can use this relationship in our original equations.

[tex]\[ a + b + c = 8 \implies a + b + a = 8 \implies 2a + b = 8 \][/tex]

[tex]\[ b = 3(a + c) \implies b = 3(2a) \implies b = 6a \][/tex]

5. Substitute [tex]\( b = 6a \)[/tex] into [tex]\( 2a + b = 8 \)[/tex]:
[tex]\[ 2a + 6a = 8 \implies 8a = 8 \implies a = 1 \][/tex]

6. Determine values of [tex]\( b \)[/tex] and [tex]\( c \)[/tex]:
Since [tex]\( a = c \)[/tex], we have [tex]\( a = 1 \)[/tex] and [tex]\( c = 1 \)[/tex].

[tex]\[ b = 6a \implies b = 6 \times 1 = 6 \][/tex]

So the digits of the number are [tex]\( a = 1 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = 1 \)[/tex]. Therefore, the three-digit number is 161.

### Conclusion
The three-digit number that satisfies all the given conditions is:
[tex]\[ \boxed{161} \][/tex]