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

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

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New post 26 Aug 2011, 06:23
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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
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Functions  [#permalink]

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New post 26 Aug 2011, 07:01
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

What's the concept behind this? Can someone please help?

Apologies - I have posted in the wrong forum before.
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Re: Functions  [#permalink]

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New post 26 Aug 2011, 09:43
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

What's the concept behind this? Can someone please help?

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.
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Re: Functions  [#permalink]

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New post 26 Aug 2011, 11:27
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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Re: Functions  [#permalink]

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New post 26 Aug 2011, 23:42
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

What's the concept behind this? Can someone please help?


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.


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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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Re: Functions  [#permalink]

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New post 27 Aug 2011, 02:37
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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Re: Functions  [#permalink]

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New post 27 Aug 2011, 08:07
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

What's the concept behind this? Can someone please help?


@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
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Re: Functions  [#permalink]

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New post 27 Aug 2011, 15:23
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

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

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New post 20 Dec 2014, 03:46
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

What's the concept behind this? Can someone please help?


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

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New post 21 Dec 2014, 21:25
adymehta29 wrote:
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

What's the concept behind this? Can someone please help?


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

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