Let's define the digits of the three-digit number:
- Let [tex]\( h \)[/tex] be the hundreds digit.
- Let [tex]\( t \)[/tex] be the tens digit.
- Let [tex]\( u \)[/tex] be the units (ones) digit.
Given the following conditions:
1. The tens digit [tex]\( t \)[/tex] is the same as the hundreds digit [tex]\( h \)[/tex], thus [tex]\( t = h \)[/tex].
2. The ones digit [tex]\( u \)[/tex] is equal to [tex]\(\frac{1}{3}\)[/tex] of the sum of the hundreds digit [tex]\( h \)[/tex] and the tens digit [tex]\( t \)[/tex]. Hence, [tex]\( u = \frac{1}{3}(h + t) \)[/tex].
3. The sum of all its digits is equal to 24, i.e., [tex]\( h + t + u = 24 \)[/tex].
Step-by-step solution:
1. Based on the first condition, we have [tex]\( t = h \)[/tex].
2. Substituting [tex]\( t = h \)[/tex] into the second condition gives:
[tex]\[
u = \frac{1}{3}(h + h) = \frac{1}{3}(2h) = \frac{2h}{3}
\][/tex]
3. Now, substituting [tex]\( t = h \)[/tex] and [tex]\( u = \frac{2h}{3} \)[/tex] into the third condition, we get:
[tex]\[
h + h + \frac{2h}{3} = 24
\][/tex]
Simplifying this equation:
[tex]\[
2h + \frac{2h}{3} = 24
\][/tex]
To clear the fraction, multiply every term by 3:
[tex]\[
3(2h) + 3\left(\frac{2h}{3}\right) = 3 \cdot 24
\][/tex]
[tex]\[
6h + 2h = 72
\][/tex]
[tex]\[
8h = 72
\][/tex]
Solving for [tex]\( h \)[/tex]:
[tex]\[
h = \frac{72}{8} = 9
\][/tex]
4. Since [tex]\( t = h \)[/tex], we have [tex]\( t = 9 \)[/tex].
5. Using [tex]\( h = 9 \)[/tex], find [tex]\( u \)[/tex]:
[tex]\[
u = \frac{2h}{3} = \frac{2 \cdot 9}{3} = \frac{18}{3} = 6
\][/tex]
So, the three-digit number is made up of digits:
- Hundreds digit: [tex]\( 9 \)[/tex]
- Tens digit: [tex]\( 9 \)[/tex]
- Units digit: [tex]\( 6 \)[/tex]
The number is thus [tex]\( \boxed{996} \)[/tex].
