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# For every integer n ≥ 3, the function g(n) is defined as the

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Intern
Joined: 15 Nov 2009
Posts: 39
For every integer n ≥ 3, the function g(n) is defined as the  [#permalink]

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26 Aug 2011, 05:23
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Difficulty:

15% (low)

Question Stats:

80% (01:07) correct 20% (01:17) wrong based on 176 sessions

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For every integer n ≥ 3, the function g(n) is defined as the product of all the odd integers from 1 to n, inclusive. What is the value of g(100) – g(99)?

A. 0
B. 1
C. 3
D. 99
E. 100
Intern
Joined: 15 Nov 2009
Posts: 39

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26 Aug 2011, 06:01
For every integer n ≥ 3, the function g(n) is defined as the product of all the odd integers from 1 to n, inclusive. What is the value of g(100) – g(99)?

1. 0
2. 1
3. 3
4. 99
5. 100

Apologies - I have posted in the wrong forum before.
Manager
Status: Quant 50+?
Joined: 02 Feb 2011
Posts: 97
Concentration: Strategy, Finance
Schools: Tuck '16, Darden '16

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26 Aug 2011, 08:43
gsaxena26 wrote:
For every integer n ≥ 3, the function g(n) is defined as the product of all the odd integers from 1 to n, inclusive. What is the value of g(100) – g(99)?

1. 0
2. 1
3. 3
4. 99
5. 100

Apologies - I have posted in the wrong forum before.

Simple concept, g(100) = g(99) because 100 is not an odd number, thus g(100) is 3X5X7....X99. The 100 does nothing.

If you didn't see this right away you could kind of cheat.

G(3) = 3, G(5) = 15, G(7) = 105, G(9) = 945.... As you can see there is no real pattern and hence way to hard to actually determine G(99) easily, therefore you will start looking for the trick.
SVP
Joined: 24 Jul 2011
Posts: 1525
GMAT 1: 780 Q51 V48
GRE 1: Q800 V740

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26 Aug 2011, 10:27
The odd integers from 1 to 100, inclusive = the odd integers from 1 to 99, inclusive
[because 100 is even]

Therefore g(100) = g(99)
=> g(100) - g(99) = 0

(A)
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26 Aug 2011, 22:42
1
gsaxena26 wrote:
For every integer n ≥ 3, the function g(n) is defined as the product of all the odd integers from 1 to n, inclusive. What is the value of g(100) – g(99)?

1. 0
2. 1
3. 3
4. 99
5. 100

Just understanding what stem is trying to convey:

for every integer $$n \ge 3$$, g(n)=Product of odd integers from 1 to n:

g(3): Here, n=3
So, g(3)=1*3(Product of ODD integers from 1 to n i.e. 1 to 3)

g(4): Here, n=4
So, g(4)=1*3 (Product of ODD integers from 1 to n i.e. 1 to 4)

Likewise,
g(99): Here, n=99
So, g(99)=1*3*5*7*....*95*97*99 (Product of ODD integers from 1 to n i.e. 1 to 99)

g(100): Here, n=100
So, g(100)=1*3*5*7*....*95*97*99 (Product of ODD integers from 1 to n i.e. 1 to 100)

You see g(99) and g(100) are same. Thus the difference must be 0.

Ans: "A"
********************************

BTW: you have posted the question in wrong forum. No PS or DS question should be posted in Math forum.

 ! Please post PS questions in the PS sub-forum: gmat-problem-solving-ps-140/Please post DS questions in the DS sub-forum: gmat-data-sufficiency-ds-141/No posting of PS/DS questions is allowed in the main Math forum.

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Joined: 29 Jun 2010
Posts: 105
WE: Information Technology (Consulting)

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27 Aug 2011, 01:37
1
For every integer n ≥ 3, the function g(n) is defined as the product of all the odd integers from 1 to n, inclusive. What is the value of g(100) – g(99)?

1. 0
2. 1
3. 3
4. 99
5. 100

g(n) is defined as product of all odd integers from 1 to n.

Lets consider a case :

let n =7 ;then g(n)=3*5*7 =75
let n =8 ;then g(n)=3*5*7 =75

There is no change in the final value of both the functions .This is because there is no odd number between 7 and 8 and hence the value is same.

The same is the case with 99 and 100. i.e. g(100) will be same as g(99)

=> g(100)-g(99) = 0 which is the answer choice A.
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Thanks,
GC24

Please click Kudos ,if my post helped you

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8789
Location: Pune, India

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27 Aug 2011, 07:07
1
1
gsaxena26 wrote:
For every integer n ≥ 3, the function g(n) is defined as the product of all the odd integers from 1 to n, inclusive. What is the value of g(100) – g(99)?

1. 0
2. 1
3. 3
4. 99
5. 100

@gsaxena26: As per your request, here is my solution.

First, we will first try and understand what the function g(n) is. g(n) is product of all odd integers upto and including n. Let me take some examples to understand this.

g(3) = 1*3 (Odd integers upto and including 3)
g(4) = 1*3 (Odd integers upto and including 4)
g(5) = 1*3*5 (Odd integers upto and including 5)
g(6) = 1*3*5 (Odd integers upto and including 6)
etc

Ok. Now we have to deal with g(99) and g(100). Let's get a sense of what they are.

g(99) = 1*3*5*7*9*...*99
g(100) = 1*3*5*7*9*...*99
They are going to be the same. Hence their difference will be 0.

Takeaway: If it is hard to understand what the question is saying, try and take some examples to understand it. Here, we tried to figure out what g(n) is using g(3), g(4) etc
_________________

Karishma
Veritas Prep GMAT Instructor

Director
Joined: 01 Feb 2011
Posts: 658

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27 Aug 2011, 14:23
for n>=3 , g(n) = product of all odd integers 1 to n.

g(100) = 1*3*5.....99

g(99) = 1*3*5*.....99

=> g(100)-g(99) = 0

Manager
Joined: 12 May 2013
Posts: 63
Re: For every integer n ≥ 3, the function g(n) is defined as the  [#permalink]

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20 Dec 2014, 02:46
VeritasPrepKarishma wrote:
gsaxena26 wrote:
For every integer n ≥ 3, the function g(n) is defined as the product of all the odd integers from 1 to n, inclusive. What is the value of g(100) – g(99)?

1. 0
2. 1
3. 3
4. 99
5. 100

gsaxena26: As per your request, here is my solution.

First, we will first try and understand what the function g(n) is. g(n) is product of all odd integers upto and including n. Let me take some examples to understand this.

g(3) = 1*3 (Odd integers upto and including 3)
g(4) = 1*3 (Odd integers upto and including 4)
g(5) = 1*3*5 (Odd integers upto and including 5)
g(6) = 1*3*5 (Odd integers upto and including 6)
etc

Ok. Now we have to deal with g(99) and g(100). Let's get a sense of what they are.

g(99) = 1*3*5*7*9*...*99
g(100) = 1*3*5*7*9*...*99
They are going to be the same. Hence their difference will be 0.

Takeaway: If it is hard to understand what the question is saying, try and take some examples to understand it. Here, we tried to figure out what g(n) is using g(3), g(4) etc

Hey Karishma,
i got this question correct.
But i am curious to know what if the question were g(102) – g(99) ?
how would you solve the question ?
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8789
Location: Pune, India
Re: For every integer n ≥ 3, the function g(n) is defined as the  [#permalink]

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21 Dec 2014, 20:25
VeritasPrepKarishma wrote:
gsaxena26 wrote:
For every integer n ≥ 3, the function g(n) is defined as the product of all the odd integers from 1 to n, inclusive. What is the value of g(100) – g(99)?

1. 0
2. 1
3. 3
4. 99
5. 100

gsaxena26: As per your request, here is my solution.

First, we will first try and understand what the function g(n) is. g(n) is product of all odd integers upto and including n. Let me take some examples to understand this.

g(3) = 1*3 (Odd integers upto and including 3)
g(4) = 1*3 (Odd integers upto and including 4)
g(5) = 1*3*5 (Odd integers upto and including 5)
g(6) = 1*3*5 (Odd integers upto and including 6)
etc

Ok. Now we have to deal with g(99) and g(100). Let's get a sense of what they are.

g(99) = 1*3*5*7*9*...*99
g(100) = 1*3*5*7*9*...*99
They are going to be the same. Hence their difference will be 0.

Takeaway: If it is hard to understand what the question is saying, try and take some examples to understand it. Here, we tried to figure out what g(n) is using g(3), g(4) etc

Hey Karishma,
i got this question correct.
But i am curious to know what if the question were g(102) – g(99) ?
how would you solve the question ?

The logic remains the same:

g(102) = 1*3*5*7*9*...*99*101
g(99) = 1*3*5*7*9*...*99

g(102) - g(99) = 1*3*5*7*9*...*99*101 - 1*3*5*7*9*...*99 = 1*3*5*7*9*...*99*(101 - 1) = 1*3*5*7*9*...*99*100
_________________

Karishma
Veritas Prep GMAT Instructor

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Posts: 9421
Re: For every integer n ≥ 3, the function g(n) is defined as the  [#permalink]

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16 Apr 2018, 23:24
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