With these set of numbers, give me the possible combinations of numbers that would total in two groups. The first group needs to be a sum of 517, and the second group 396: 48, 40, 72, 72, 80, 68, 79, 58, 88, 74, 100, 36, 98, 92

To solve this, we need to partition the provided set of numbers into two groups where the sum of the first group is 517 and the sum of the second group is 396.

The given numbers are:

48, 40, 72, 72, 80, 68, 79, 58, 88, 74, 100, 36, 98, 92

### Total Sum
The total sum of all the numbers is:

\[
48 + 40 + 72 + 72 + 80 + 68 + 79 + 58 + 88 + 74 + 100 + 36 + 98 + 92 = 1005
\]

Since the total sum is 1005, partitioning it into two groups where one sums to 517 and the other sums to 396 is possible because:

\[
517 + 396 = 913
\]

There is a discrepancy of 1005 - 913 = 92. This means we might need to eliminate or exclude one or more numbers.

I'll now look for possible combinations that achieve the sums of 517 and 396, and verify them.

Let me compute this.It appears that no exact combination of numbers gives a sum of 517 and 396 while using all the provided numbers. This may be because the total sum of the numbers (1005) leaves a discrepancy when trying to split it evenly.

Let me explore alternate combinations, where we might need to exclude a number or slightly adjust. Would you like me to try excluding one or more numbers to see if a partition becomes possible?

With these set of numbers, give me the possible combinations of numbers that would total in two groups. The first group needs to be a sum of 517, and the second group 396: 48, 40, 72, 72, 80, 68, 79, 58, 88, 74, 100, 36, 98, 92.

Explore how to partition a given set of numbers to form two groups with sums of 517 and 396. This problem highlights concepts in combinatorics and number theory, with step-by-step exploration of the total sum, exclusions, and possible combinations. Suitable for students in Grades 10-12.

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