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# For every positive even integer n, the function h(n) is

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25 Feb 2010, 08:25
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25 Feb 2010, 08:36
Expert's post
ThePower wrote:

GMAT won't ask you to factor such a huge numbers as $$2^{50}*50!+1$$. The stem of the question is enough to answer the question without factoring it. The answer is E (more than 40).
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25 Feb 2010, 09:51
Bunuel wrote:
msv3763 wrote:
For every positive even integer n, the function h(n) is defnied to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100)+1, then p is

Any help on how to get the correct answer? Thanks!

Welcome to the Gmat Club. Below is the solution for your question:

$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$

Now, two numbers $$h(100)=2^{50}*50!$$ and $$h(100)+1=2^{50}*50!+1$$ are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

As $$h(100)=2^{50}*50!$$ has all numbers from 1 to 50 as its factors, according to above $$h(100)+1=2^{50}*50!+1$$ won't have ANY factor from 1 to 50. Hence $$p$$ ($$>1$$), the smallest factor of $$h(100)+1$$ will be more than 50.

Hope it helps.

P.S. Can you please: post one question per topic, tag the questions you post, and also post the whole questions with answer choices.

Good explanation.
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26 Feb 2010, 09:48
Bunuel wrote:
msv3763 wrote:
For every positive even integer n, the function h(n) is defnied to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100)+1, then p is

Any help on how to get the correct answer? Thanks!

Welcome to the Gmat Club. Below is the solution for your question:

$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$

Now, two numbers $$h(100)=2^{50}*50!$$ and $$h(100)+1=2^{50}*50!+1$$ are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

As $$h(100)=2^{50}*50!$$ has all numbers from 1 to 50 as its factors, according to above $$h(100)+1=2^{50}*50!+1$$ won't have ANY factor from 1 to 50. Hence $$p$$ ($$>1$$), the smallest factor of $$h(100)+1$$ will be more than 50.

Hope it helps.

P.S. Can you please: post one question per topic, tag the questions you post, and also post the whole questions with answer choices.

Very nice way of solving the problem. Bunuel, you make the problem look easy.
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26 Feb 2010, 12:22
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ThePower wrote:

There's no way to prove it without a computer; factoring large integers is one of the most time-consuming mathematical operations (unless they have very obvious factors). The fact that it is so time consuming to factor large numbers is actually the basis of internet encryption systems; you can hack the encryption, but you'd need 10,000 years of computing time to break down the large numbers involved. So Number Theory is the reason you can do banking, shopping or private email on the web.
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27 Feb 2010, 23:15
an after thought.......

Cannot understand why according to above h (100)+1 won't have ANY factor from 1 to 50. Let's take the example of 20 & 21; is it true that 21 doesn't have any factor till 20? They won't have common factors like factors for 20 are 2 & 5 while of 21 are 7 & 3, but 21 does have factors less than 20.

Pl enlighten.
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27 Feb 2010, 23:36
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ExecMBA2010 wrote:
an after thought.......

Cannot understand why according to above h (100)+1 won't have ANY factor from 1 to 50. Let's take the example of 20 & 21; is it true that 21 doesn't have any factor till 20? They won't have common factors like factors for 20 are 2 & 5 while of 21 are 7 & 3, but 21 does have factors less than 20.

Pl enlighten.

Not sure I understood your question correctly.

Anyway: given that $$p$$ is the smallest prime factor of $$h(100)+1=2^{50}*50!+1$$. The point here is that consecutive integers do not share ANY common factor but 1. Two numbers $$h(100)=2^{50}*50!$$ and $$h(100)+1=2^{50}*50!+1$$ are consecutive integers, hence these two integers do not share any common factor but 1. $$h(100)=2^{50}*50!$$ naturally has all factors from 2 to 50, which means that $$h(100)+1=2^{50}*50!+1$$ does not have any of these factors. So, $$p$$, which is the prime factor of $$h(100)+1=2^{50}*50!+1$$ must be more than 50.
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28 Feb 2010, 07:18
Bunuel wrote:
Not sure I understood your question correctly.

Anyway: given that $$p$$ is the smallest prime factor of $$h(100)+1=2^{50}*50!+1$$. The point here is that consecutive integers do not share ANY common factor but 1. Two numbers $$h(100)=2^{50}*50!$$ and $$h(100)+1=2^{50}*50!+1$$ are consecutive integers, hence these two integers do not share any common factor but 1. $$h(100)=2^{50}*50!$$ naturally has all factors from 2 to 50, which means that $$h(100)+1=2^{50}*50!+1$$ does not have any of these factors. So, $$p$$, which is the prime factor of $$h(100)+1=2^{50}*50!+1$$ must be more than 50.

I have got it now. Since h(100) has all the numbers from 2 to 50 as its factor h(100)+1 cannot have any of the numbers between 2 & 50 as its factor. Hence the lowest factor of h(100)+1 is either 1 or a number above 50.

Thanks.
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14 Apr 2010, 06:58
#1 For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) +1, the p is?
A. btw 2 and 20
B. btw 10 and 20
C. btw 20 and 30
D. btw 30 and 40
E. greater than 40

#2 Pumps A, B, and C operate at their respective constant rates. Pumps A and B, operating simultaneously, can fill a certain tank in 6.5 hours; pumps A and C, operating simultaneously, can fill the tank in 3.2 hours; and pumps B and C, operating simultaneously, can fill the tank in 2 hours. How many hours does it take pumps A,B, and C operating simultaneously, to fill the tank?

#3 Last year certain bond with a face value of $5,000 yielded 8 percent of its face value in interest. If that interest was approximately 6.5 percent of the bond's selling price, approximately what was the bond's selling price?$4,063
$5,325$5,351
$6,000$6,154
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14 Apr 2010, 08:28
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Quote:
#1 For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) +1, the p is?
A. btw 2 and 20
B. btw 10 and 20
C. btw 20 and 30
D. btw 30 and 40
E. greater than 40

Bit Tricky..
The function described will be in the form h(n)= 2^(n/2)*(n/2)!
h(100)+1= [(2^50)*50! +1]

By POE we can eliminate ABCD(because any prime number below 50 is a factor of 50!)

I feel the ans is E
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14 Apr 2010, 08:39
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CC2HBS wrote:
#3 Last year certain bond with a face value of $5,000 yielded 8 percent of its face value in interest. If that interest was approximately 6.5 percent of the bond's selling price, approximately what was the bond's selling price?$4,063
$5,325$5,351
$6,000$6,154

x- Selling Price
8%5000 = 6.5%x
x = (8 * 5000)/6.5
= 6154 hence E.
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14 Apr 2010, 08:51
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I tried to explain Q2 1 it is very hard to explain 2 it has a mammoth calculation

but can brief::: the answer will be (1/6.5 +1/3.2 + 1/2)*1/2...
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14 Apr 2010, 08:55
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14 Apr 2010, 09:00
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Expert's post
#1 For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) +1, the p is?
A. btw 2 and 20
B. btw 10 and 20
C. btw 20 and 30
D. btw 30 and 40
E. greater than 40

$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$

Now, two numbers $$h(100)=2^{50}*50!$$ and $$h(100)+1=2^{50}*50!+1$$ are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

As $$h(100)=2^{50}*50!$$ has all numbers from 1 to 50 as its factors, according to above $$h(100)+1=2^{50}*50!+1$$ won't have ANY factor from 1 to 50. Hence $$p$$ ($$>1$$), the smallest factor of $$h(100)+1$$ will be more than 50.

#2 Pumps A, B, and C operate at their respective constant rates. Pumps A and B, operating simultaneously, can fill a certain tank in 6.5 hours; pumps A and C, operating simultaneously, can fill the tank in 3.2 hours; and pumps B and C, operating simultaneously, can fill the tank in 2 hours. How many hours does it take pumps A,B, and C operating simultaneously, to fill the tank?

I think there is a typos in stem: times should be 6/5 (instead of 6.5) and 3/2 (instead of 3.2).

THEORY
If:
Time needed for A to complete the job =A hours;
Time needed for B to complete the job =B hours;
Time needed for C to complete the job =C hours;
...
Time needed for N to complete the job =N hours;

Then if time needed for all of them working simultaneously to complete the job is $$T$$, then: $$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+..+\frac{1}{N}=\frac{1}{T}$$ (General formula).

For two and three entities (workers, pumps, ...):

General formula for calculating the time needed for two workers A and B working simultaneously to complete one job:

Given that $$a$$ and $$b$$ are the respective individual times needed for $$A$$ and $$B$$ workers (pumps, ...) to complete the job, then time needed for $$A$$ and $$B$$ working simultaneously to complete the job equals to $$T_{(A&B)}=\frac{a*b}{a+b}$$ hours, which is reciprocal of the sum of their respective rates ($$\frac{1}{a}+\frac{1}{b}=\frac{1}{t}$$).

General formula for calculating the time needed for three A, B and C workers working simultaneously to complete one job:

$$T_{(A&B&C)}=\frac{a*b*c}{ab+ac+bc}$$ hours.

Also for rate problems it's good to know that:

TIME to complete one job=Reciprocal of rate. eg 6 hours needed to complete one job (time) --> 1/6 of the job done in 1 hour (rate).

Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance.

Time*Rate=Distance
Time*Rate=Job

Back to our original question.
We have:
$$\frac{1}{A}+\frac{1}{B}=\frac{1}{t_{a&b}}=\frac{5}{6}$$
$$\frac{1}{A}+\frac{1}{C}=\frac{1}{t_{a&c}}=\frac{2}{3}$$
$$\frac{1}{B}+\frac{1}{C}=\frac{1}{t_{b&c}}=\frac{1}{2}$$

Question: $$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{T}$$, $$T=?$$

Sum the first three equations: $$2(\frac{1}{A}+\frac{1}{B}+\frac{1}{C})=\frac{5}{6}+\frac{2}{3}+\frac{1}{2}=2$$ --> $$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=1=\frac{1}{T}$$ --> $$T=1$$.
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14 Apr 2010, 09:07
RaviChandra wrote:

first of all thx for all the kind explanations!! it is still confused to understand whole process but i guess i started to get a grip!
#1. greater than 40
#2. 1
#3. 6,154
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14 Apr 2010, 20:00
Quote:
Pumps A, B, and C operate at their respective constant rates. Pumps A and B, operating simultaneously, can fill a certain tank in 6.5 hours; pumps A and C, operating simultaneously, can fill the tank in 3.2 hours; and pumps B and C, operating simultaneously, can fill the tank in 2 hours. How many hours does it take pumps A,B, and C operating simultaneously, to fill the tank?

I think there is a typos in stem: times should be 6/5 (instead of 6.5) and 3/2 (instead of 3.2)

How could you identify that there is a typo dude... ur simply great
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03 May 2010, 13:02
For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If p is the smallest prime factior of h (100) + 1, the p is:

* between 2 and 10
* between 10 & 20
* between 20 & 30
* between 30 & 40
* greater than 40

If n is a positive integer and the product of all integers from 1 to n inclusive is a multiple of 990, what is the least possible value of n?

*10
*11
*12
*13
*14
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03 May 2010, 13:05
If n is a positive integer and the product of all integers from 1 to n inclusive is a multiple of 990, what is the least possible value of n?

*10
*11
*12
*13
*14

Sol - Product of n numbers = $$1*2*3.... n = n ! = 990 * K = 9* 10 *11 * K$$

The least possible value of n will be 11, as 11 is a prime number which is there in the product of the n numbers.
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14 May 2010, 04:48
I will tell a more simple and direct solution for this question.

Remember the Wilson's Theorem

p divides (p-1)!+1 if p is prime. The converse is also true.

For instance, 7 is prime, and 7 divides (6!)+1=721

For h(100) = 100!

h(100)+1 = (100!+1)

Which (keeping the Wilson's theorem in mind) can be rewritten as

h(101-1)+1 = [ (101-1)! + 1 ]

101 is the prime which shall divide h(100)+1

and 101 is greater than 40, and choose this answer choice.

Magic lies in the Wilson's theorem, but if you remember the above mentioned method, it will help you in solving any similar or related question without any problem at all.
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Re: GMAT Prep Question on Integers [#permalink]

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15 May 2010, 21:53
Great Explanation!
Re: GMAT Prep Question on Integers   [#permalink] 15 May 2010, 21:53

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