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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Very nice way of solving the problem. Bunuel, you make the problem look easy.

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.

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.

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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One Final Try.......

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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One Final Try.......

#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.

Answer: E.


#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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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!


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How could you identify that there is a typo dude... ur simply great
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One Final Try.......

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.

Great Explanation!