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So S1 is 0, so S3 = S2 + 0, then eventually, S4 = S3 + S2, which is 2*S2!

So similarly, if you keep on substituting values, you’ll reach S7 = 5*S2 + 3*S2, and since 7=n-1, n=8
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^^^

Told in the question that Sn-1 = 8S2 , and we calculate that S7 = 8S2, set them equal to one another and n = 8
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Susan invited 13 of her friends for her birthday party and created return gift hampers comprising one each of $3, $4, and $5 gift certificates. One of her friends did not turn up and Susan decided to rework her gift hampers such that each of the 12 friends who turned up got $13 worth gift certificates. How many gift hampers did not contain $5 gift certificates in the new configuration?
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Anyone tried scheduling the gmat classic exam at a test centre in the month of January?
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Md.RakibulHasan
Susan invited 13 of her friends for her birthday party and created return gift hampers comprising one each of $3, $4, and $5 gift certificates. One of her friends did not turn up and Susan decided to rework her gift hampers such that each of the 12 friends who turned up got $13 worth gift certificates. How many gift hampers did not contain $5 gift certificates in the new configuration?
3 sets

we have 3,4,5 for 13 people which is redistributed among 12 ppl (so now sum for each person is 13). We can make this by 5 5 3; 4 4 5; 3 3 3 4. Now Qn asks to find how many sets dont have a 5 i.e. no of sets with 3334 combination. So far was the hard bit, now just solve (again dont manually solve as then the 2nd bit will be equally hard). Since you have 3334 we know the number of sets of these will have to be less than 5 (Why? coz total is 13 pieces available). Lets start with 4, means only 1 set of 553 will be possible wont work as we need to use up all 5’s. Try with 3, we get (5 5 3) *4 sets, (4 4 5)*5sets and (3 3 3 4)*3 sets. Total sets are 4+5+3=12 sets. Voila :)

Damn typing this took longer than solving it
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Thank you so much for your detailed explanation. I was stuck in for few days with this math’s written method. Now It’s clear to me.
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Happy to hear that it was useful to you :)
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Hi, I went through some Q’s from the Manhattan Prep books and also been doing some Q’s from the OG book. Noticed that there’s no Phytogoras Q’s in the OG book so far. Understand the Geometry is not being tested anymore. But concerned Phytogoras maybe tested via graphs.
Q: Is Phytogoras Theorem going to be tested in FE edition?
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Malar95
Hi, I went through some Q’s from the Manhattan Prep books and also been doing some Q’s from the OG book. Noticed that there’s no Phytogoras Q’s in the OG book so far. Understand the Geometry is not being tested anymore. But concerned Phytogoras maybe tested via graphs. Q: Is Phytogoras Theorem going to be tested in FE edition?
What do You mean by "Geometrie is not tested anymore" in the GMAT?
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dominicusbenacus
What do You mean by "Geometrie is not tested anymore" in the GMAT?
Geometry is not part of FE Quant.
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You mean GMAT Focus Edition aka FE Quant?
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Back to my first Q. Is Phytogoras part of new FE Quant?
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Malar95
Back to my first Q. Is Phytogoras part of new FE Quant?
I haven’t seen any question in Quant section, while taking mocks, being based on Pythagoras. But don’t know for sure if it ever be asked in some way or another.
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Help me solve this question - Let P be the set of all seven digit numbers formed using the digits 1,2,3,4,5,6 and 7 without repetition such that the exactly three odd positions are occupied by the odd numbers. What is the sum of all the unit’s digits of the members of P?
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srush283
Help me solve this question - Let P be the set of all seven digit numbers formed using the digits 1,2,3,4,5,6 and 7 without repetition such that the exactly three odd positions are occupied by the odd numbers. What is the sum of all the unit’s digits of the members of P?
What is the source of this question? I have a rough idea of how to calculate this, but it is a monster that would take 10-15 minutes at best
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how many different numbers can be obtained as the product of exactly 3 different numbers in the set (2, 3, 5, 7, 11)

6

10, 12, 15, 20

I thought 5C3 since the set is composed only by prime numbers so 3 multiplied numbers cannot have the same result. Is it correct?
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