To solve the problem of filling three boxes such that the sum of the numbers in the boxes is 30, using only the numbers from the set [tex]\(\{1, 3, 5, 7, 9, 11, 13, 15\}\)[/tex] and allowing repetition, we need to examine all possible combinations of these numbers.
First, let's define the set of possible numbers:
[tex]\[
\{1, 3, 5, 7, 9, 11, 13, 15\}
\][/tex]
We need to find all combinations of three numbers from this set such that their sum equals 30.
[tex]\[a + b + c = 30\][/tex]
Let's investigate the possible combinations:
1. Start by checking the largest number, 15. If we use 15, then the equation reduces to [tex]\(b + c = 15\)[/tex]. However, both [tex]\(b\)[/tex] and [tex]\(c\)[/tex] must also be in the set of possible numbers.
- If [tex]\(b = 15\)[/tex], then [tex]\(c = 0\)[/tex] which is not in the set.
- If [tex]\(b = 13\)[/tex], then [tex]\(c = 2\)[/tex] which is not in the set.
- If [tex]\(b = 11\)[/tex], then [tex]\(c = 4\)[/tex] which is not in the set.
- If [tex]\(b = 9\)[/tex], then [tex]\(c = 6\)[/tex] which is not in the set.
- If [tex]\(b = 7\)[/tex], then [tex]\(c = 8\)[/tex] which is not in the set.
- If [tex]\(b = 5\)[/tex], then [tex]\(c = 10\)[/tex] which is not in the set.
- If [tex]\(b = 3\)[/tex], then [tex]\(c = 12\)[/tex] which is not in the set.
- If [tex]\(b = 1\)[/tex], then [tex]\(c = 14\)[/tex] which is not in the set.
2. Next, let's consider the second-largest number, 13:
- If we substitute [tex]\(a = 13\)[/tex], then [tex]\(b + c = 17\)[/tex].
- By examining all pairs in the set, we find that 17 cannot be expressed as a sum of two numbers from the set.
3. Similarly, let's check for 11:
- Substituting [tex]\(a = 11\)[/tex], we need [tex]\(b + c = 19\)[/tex].
- Again, no combination of numbers in the set adds up to 19.
4. Repeat this check for 9, 7, 5, 3:
- For [tex]\(a = 9\)[/tex], [tex]\(b + c = 21\)[/tex], no valid combinations.
- For [tex]\(a = 7\)[/tex], [tex]\(b + c = 23\)[/tex], no valid combinations.
- For [tex]\(a = 5\)[/tex], [tex]\(b + c = 25\)[/tex], no valid combinations.
- For [tex]\(a = 3\)[/tex], [tex]\(b + c = 27\)[/tex], no valid combinations.
- For [tex]\(a = 1\)[/tex], [tex]\(b + c = 29\)[/tex], no valid combinations.
After checking all these possible sets of combinations, we find that there are no three numbers in the given set that can add up to 30. Therefore, there are no valid combinations of three numbers from the set [tex]\(\{1, 3, 5, 7, 9, 11, 13, 15\}\)[/tex] that sum up to 30.
Hence, the solution to the problem is that there are no valid combinations of three numbers from the given set that sum to 30.
