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S98-17

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S98-17 [#permalink]

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New post 16 Sep 2014, 01:52
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Difficulty:

  95% (hard)

Question Stats:

19% (00:51) correct 81% (01:39) wrong based on 63 sessions

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Kudos [?]: 135286 [0], given: 12679

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Re S98-17 [#permalink]

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New post 16 Sep 2014, 01:52
Official Solution:

How many different sets of positive square integers, each greater than 1, add up to 75?

A. 1
B. 4
C. 7
D. 11
E. 13


First, lay out the possible numbers you can use in the sum. The positive square integers greater than 1 but less than 75 are 4, 9, 16, 25, 36, 49, and 64.

Now, let's create a possible set and then see whether we can adjust it. It's fairly obvious that three 25's add up to 75, so our first set is {25, 25, 25}.

We might now recall the most famous example of the Pythagorean Theorem: \(9 + 16 = 25\). So we can swap out, successively, a 25 and replace it with a 9 and a 16. With sets, order does not matter, so we get three more possible sets:

{25, 25, 9, 16}

{25, 9, 16, 9, 16}

{9, 16, 9, 16, 9, 16}

This gives us 4 sets so far. However, we can now swap out 16's for four 4's. We can do so as follows.

One possible swap for the set with one 16:

{25, 25, 9, 4, 4, 4, 4}

Two possible swaps for the set with two 16's:

{25, 9, 16, 9, 4, 4, 4, 4}

{25, 9, 4, 4, 4, 4, 9, 4, 4, 4, 4}

And three possible swaps for the set with three 16's:

{9, 16, 9, 16, 9, 4, 4, 4, 4}

{9, 16, 9, 4, 4, 4, 4, 9, 4, 4, 4, 4}

{9, 4, 4, 4, 4, 9, 4, 4, 4, 4, 9, 4, 4, 4, 4}

Before going to the larger squares, we should glance over our list and see whether we can do any swaps within the sets we've already created, using only squares equal to 25 or less. The only swap we can do is in the last set: we can swap out nine 4's and replace them with four 9's:

{9, 4, 4, 4, 4, 9, 4, 4, 4, 4, 9, 4, 4, 4, 4} = three 9's and twelve 4's

becomes

{9, 9, 9, 9, 9, 9, 9, 4, 4, 4} = seven 9's and three 4's

We are now at a total of 11 sets, having exhausted the possibilities that only involve the squares equal to 25 or less. Are there any sets that involve larger squares?

We can quickly check:

64 can't be in the set, because the leftover (11) can't be formed from the sum of 9's and/or 4's.

49 CAN be in the set. The leftover (26) can be formed by the sum of two 9's, and two 4's, so we get

{49, 9, 9, 4, 4}

36 CAN be in the set. The leftover (39) can be written as the sum of three 9's and three 4's, so we get

{36, 9, 9, 9, 4, 4, 4}

Thus, the total number of different sets is 13.


Answer: E
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Kudos [?]: 135286 [0], given: 12679

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Re: S98-17 [#permalink]

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New post 23 Mar 2015, 01:53
this should be 700+ difficulty as it is taking so much time....

Any shortcut for this?

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Re: S98-17 [#permalink]

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New post 23 Mar 2015, 03:27

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Re: S98-17 [#permalink]

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New post 29 Aug 2016, 19:24
This is not very hard, but since there is no formula to get to the answer, it's going to take a long time. This is definitely 750+ question.

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Re: S98-17 [#permalink]

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New post 29 Aug 2016, 19:27
If this question is unlikely to show up in an actual GMAT test, what is the reason it is asked here in GMAT club test?

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Re: S98-17 [#permalink]

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New post 18 Aug 2017, 00:30
I hope there must be some logic to this question. These kind of questions need to be skipped after 60 second max.

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Re: S98-17   [#permalink] 18 Aug 2017, 00:30
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