M04-27

Question

If  S  is the set of all fractions defined by the formula  \frac{n}{n+1} , where  n  is a positive integer less than 20, what is the product of all the fractions that are in  S ?

A.  \frac{1}{20}

B.  \frac{1}{21}

C.  \frac{1}{2}

D.  \frac{19}{40}

E.  \frac{19}{20}

Bunuel · Answer

Official Solution:

If  S  is the set of all fractions defined by the formula  \frac{n}{n+1} , where  n  is a positive integer less than 20, what is the product of all the fractions that are in  S ?

A.  \frac{1}{20}

B.  \frac{1}{21}

C.  \frac{1}{2}

D.  \frac{19}{40}

E.  \frac{19}{20}

\frac{1}{1+1} * \frac{2}{2+1} * \frac{3}{3+1} * \frac{4}{4+1} * ... * \frac{19}{19 +1} =  
 
 =\frac{1}{2} * \frac{2}{3} * \frac{3}{4} *\frac{4}{5}* ... * \frac{19}{20} =  
 
 =\frac{19!}{20!} =  
 
 =\frac{1}{20}

Answer: A

BobsterGMAT · Answer

dang this is a good question)

PN0593 · Answer

I am still very confused- how did you get 1/20 from 19/20?

Bunuel · Answer

It's not 19/20, it's 19!/20!. Factorial of 19 divided by factorial of 20.

djTae · Answer

The problem is asking for the “product of all the fractions that are in S,” which is how we come to think of it as a factorial.

The factorial is essentially the shortened form. If we write out the full problem, we see that because we are multiplying the fractions the denominator of each fraction in the set is cancelled by the numerator of the following fraction, except the first numerator and the last denominator.

Example:
 \frac{1}{(1+1)}  *  \frac{2}{(2+1)}  ...  \frac{19}{(19+1)}  =  \frac{1}{2}  *  \frac{2}{3}  ... *  \frac{19}{20}

Once you reduce, all that is left is the first numerator, 1, and the last denominator, 20.

Leaving us with  \frac{1}{20}

Posted from my mobile device   
Edit: Correct fraction readability per rules.

Ravixxx · Answer

A quick way (and easy) to answer this question is trying to recognize the pattern.

Two important things:
1) we must find a product of the set.
2) n/(n+1)

Start with the first 5 numbers.

1= 1/2 
2= 2/3 
3= 3/4 
4= 4/5
5= 5/6

Now spot the pattern:
1/2*2/3 = 2/6 = 1/3
1/3*3/4 (you can cross-multiply)= 3/12 = 1/4 
1/4*4/5 (you can cross-multiply) = 1/5
1/5*5/6 (you can cross-multiply) = 1/6

Have you spot the pattern? For each product you can cross multiply
1/4 —> 1/5 —> 1/6

Since last number is 1/(19+1) the final product will be 1/20.

Option A.

Posted from my mobile device

TheNightKing · Answer

Just try it in your head.

1/2 * 2/3 * 3/4 *...... * 19/20

(Be careful. n is less than 20 so max is 19 so the last fraction is 19/20)

Everything but 1/20 cancels out.

Answer A

RenanBragion · Answer

I think this is a  high-quality  question and I agree with explanation.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

UkrHurricane · Answer

While here the correct answer can be deduced based on the answer choices, why can't n be say 18? It is an integer and is less than 20.

Bunuel · Answer

The question asks to find the  product  of all the fractions that are in S.

S consists of the following fractions  \frac{1}{1+1}, \ \frac{2}{2+1}, \ \frac{3}{3+1}, \ \frac{4}{4+1}, \ ... , \ \frac{19}{19 +1}

The product of which is:

=\frac{1}{2} * \frac{2}{3} * \frac{3}{4} *\frac{4}{5}* ... * \frac{19}{20} =  
 
 =\frac{19!}{20!} =  
 
 =\frac{1}{20}

M04-27

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