Which of the following is an integer

Question

Which of the following expressions is an integer?

(A)  \sqrt{\frac{14!\cdot 15!}{2}} 
(B)  \sqrt{\frac{15!\cdot 16!}{2}} 
(C)  \sqrt{\frac{16!\cdot 17!}{2}} 
(D)  \sqrt{\frac{17!\cdot 18!}{2}} 
(E)  \sqrt{\frac{18!\cdot 19!}{2}}

dhruv23320 · Answer

We want to determine for which choice of n the expression sqrt(n!(n+1)!/2) is an integer.

Compute: sqrt(n!(n+1)!/2) = sqrt((n+1)(n!)^2/2) = n! · sqrt((n+1)/2).

Therefore, the expression is an integer if (n+1)/2 is a perfect square.

Now check each option:

(A) Not a perfect square

(B): Not a perfect square

(C): Not a perfect square

(D): A perfect square

(E): Not a perfect square

Only (D) gives (n+1)/2 = 9, a perfect square, so the whole expression is an integer.

Answer: (D)

naw · Answer

Would love to hear an expert reply. 
How can you know which ones are perfect squares without calculating it

Adit_ · Answer

Using POE we can solve this:

1st one:
We have an issue with factor 5 because:
14! has 2 5s only and 15! brings another extra 5 making it 5^3 whose root cannot give an integer. OUT.

2nd one:
We have an issue with factor 2 itself here.
16! gives 15!*16 and 16=2^4, since there is divided by 2 included, we have to remove one of the 2s, it leaves us with odd number of 2s. Again cannot give us an integer. OUT.

3rd one: 17 is a prime number and there is no other 17 in the expression, but only 1. Not an integer. OUT.

4th one:
Here 17 gets multiplied again in 18! thus that is not a problem here. Taking 5 as another factor also works as 17/5 and 18/5 results in same values. 
So far looks good no need to dive further we havent yet found an issue with this. Keep.

5th one:
Ending with prime number 19 in 19!, no other 19 in expression and only 1 of it is there. Root can't be an integer. OUT.

Our final answer has to be  option D  only.

Adit_ · Answer

You can otherwise solve using POE, you need to find just 1 reason to eliminate an option, if you find 2 or more reasons to keep the option then keep, and keep eliminating the rest until you have hit on the final answer. In this case by using the prime factorization method we can eliminate all the options except D.

naw

Would love to hear an expert reply.
How can you know which ones are perfect squares without calculating it

KarishmaB · Answer

When we have a square root, to get an integer answer, we need 2 instances of the number under the square root.

e.g.  \sqrt{4} = \sqrt{2*2} = 2  (integer)
 \sqrt{100} = \sqrt{10*10} = 10  (integer)

So what we are looking for is 2 instances of each number so that we can take it out of the root.

Each n! can be written as n(n-1)! and hence 14! * 15! =14! * 14! * 15 (two instances of 14!)
So we will take 14! out of square root. The point is whether the leftover inside the square root has 2 instances as well.

\sqrt{\frac{14!\cdot 15!}{2}} = \sqrt{\frac{14! * 14! * 15}{2}} = 14! * \sqrt{\frac{15}{2}} 
What is inside the square root now is not an integer. So Ignore

but for 
 \sqrt{\frac{17!\cdot 18!}{2}} = \sqrt{\frac{17!*17! * 18}{2}}= 17! * \sqrt{\frac{18}{2}} = 17! * \sqrt{9} = 17! * \sqrt{3*3} = 17! * 3   (Ań integer)

Answer (D)

Which of the following is an integer

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GMAT Problem Solving (PS) Questions

Which of the following is an integer

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