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i don't understand why we can conclude five subsets summing to 142, those remaining if we remove {10}, {2, 8}, {3, 7}, {4, 6}, and {2, 3, 5}, from {2, 3, 4, ..., 17}.
the calculation will be "Total subsets - no of subsets have to be removed = number of subsets summing 142". however, we just know the number of subsets that have to be removed. how to calculate the total subset?
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i don't understand why we can conclude five subsets summing to 142, those remaining if we remove {10}, {2, 8}, {3, 7}, {4, 6}, and {2, 3, 5}, from {2, 3, 4, ..., 17}.
the calculation will be "Total subsets - no of subsets have to be removed = number of subsets summing 142". however, we just know the number of subsets that have to be removed. how to calculate the total subset?
­
Removing {10} from S:
Subset: {2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17}

Removing {2, 8} from S:
Subset: {3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 16, 17}

Removing {3, 7} from S:
Subset: {2, 4, 5, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}

Removing {4, 6} from S:
Subset: {2, 3, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}

Removing {2, 3, 5} from S:
Subset: {4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}

Each of these subsets sums to 142
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Hi Bunuel,

In questions like these, I am always worried I have missed a test case. Is there a different way to solve this besides the brute force method?


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i don't understand why we can conclude five subsets summing to 142, those remaining if we remove {10}, {2, 8}, {3, 7}, {4, 6}, and {2, 3, 5}, from {2, 3, 4, ..., 17}.
the calculation will be "Total subsets - no of subsets have to be removed = number of subsets summing 142". however, we just know the number of subsets that have to be removed. how to calculate the total subset?
­
Removing {10} from S:
Subset: {2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17}

Removing {2, 8} from S:
Subset: {3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 16, 17}

Removing {3, 7} from S:
Subset: {2, 4, 5, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}

Removing {4, 6} from S:
Subset: {2, 3, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}

Removing {2, 3, 5} from S:
Subset: {4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}

Each of these subsets sums to 142
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Moh12345
Hi Bunuel,

In questions like these, I am always worried I have missed a test case. Is there a different way to solve this besides the brute force method?


Bunuel
phamminh2003
i don't understand why we can conclude five subsets summing to 142, those remaining if we remove {10}, {2, 8}, {3, 7}, {4, 6}, and {2, 3, 5}, from {2, 3, 4, ..., 17}.
the calculation will be "Total subsets - no of subsets have to be removed = number of subsets summing 142". however, we just know the number of subsets that have to be removed. how to calculate the total subset?
­
Removing {10} from S:
Subset: {2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17}

Removing {2, 8} from S:
Subset: {3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 16, 17}

Removing {3, 7} from S:
Subset: {2, 4, 5, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}

Removing {4, 6} from S:
Subset: {2, 3, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}

Removing {2, 3, 5} from S:
Subset: {4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}

Each of these subsets sums to 142
The method shown in the original solution is the most straightforward way of solving this problem. If done properly and carefully, it should not take much time either. I think confidence will come with more practice.
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While I got this right, I did not take into account the null set, shouldn't that also be added, making it 5 + 1 ?
Bunuel
Official Solution:

If the data set S = {2, 3, 4, ..., 17}, how many subsets of S have a sum of 142?

A. 2
B. 3
C. 4
D. 5
E. 6


Firstly, let's calculate the sum of the original set: \(2 + 3 + 4 + ... + 17 = \frac{2 + 17}{2} * 16 = 152\).

We need to find the number of subsets with a sum of 142, which is 10 less than the sum of S. Essentially, we need to count the subsets that total 10, because excluding them from S results in a subset summing to 142.

• The one-element subset summing to 10 is {10};

• Two-element subsets summing to 10 are {2, 8}, {3, 7}, and {4, 6}.

• The only three-element subset summing to 10 is {2, 3, 5}.

We cannot have a four-element subset summing to 10 because the smallest sum of any four elements in S is \(2 + 3 + 4 + 5 = 14\).

Therefore, there are five subsets summing to 142, those remaining if we remove {10}, {2, 8}, {3, 7}, {4, 6}, and {2, 3, 5}, from {2, 3, 4, ..., 17}.


Answer: D
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Ada10
While I got this right, I did not take into account the null set, shouldn't that also be added, making it 5 + 1 ?
Bunuel
Official Solution:

If the data set S = {2, 3, 4, ..., 17}, how many subsets of S have a sum of 142?

A. 2
B. 3
C. 4
D. 5
E. 6


Firstly, let's calculate the sum of the original set: \(2 + 3 + 4 + ... + 17 = \frac{2 + 17}{2} * 16 = 152\).

We need to find the number of subsets with a sum of 142, which is 10 less than the sum of S. Essentially, we need to count the subsets that total 10, because excluding them from S results in a subset summing to 142.

• The one-element subset summing to 10 is {10};

• Two-element subsets summing to 10 are {2, 8}, {3, 7}, and {4, 6}.

• The only three-element subset summing to 10 is {2, 3, 5}.

We cannot have a four-element subset summing to 10 because the smallest sum of any four elements in S is \(2 + 3 + 4 + 5 = 14\).

Therefore, there are five subsets summing to 142, those remaining if we remove {10}, {2, 8}, {3, 7}, {4, 6}, and {2, 3, 5}, from {2, 3, 4, ..., 17}.


Answer: D

Does an empty set have a sum of 142? Why should it be included, then?
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