Bill can dig a well in x! hours. Carlos can dig the same

To understand this, try digits.
y=5
x=5+1=6
6!=1*2*3*4*5*6
or
(5+1)5! = (5+1)*1*2*3*4*5

x-y=1
=> x=y+1

=> x! = (y+1)! = (y+1)*y!

use the property....x! = x*(x-1) *(x-2) .....1------------------1

(x-1)! = (x-1)*(x-2)*......1------------------------------------2

Using equation 1 and 2
we get x! = (x)*(x-1)!

As per stmt 1: x-y=1 +> x=y+1
thus x! = (y+1)!

Now q = (x! * y!) / (x! + y!)
=> q = ((y+1)! * y!) / ((y+1)! + y!)

Taking y! as common in both numerator and denominator

q = y!*(y+1)! / y!*(y+1 +1)
=> q = (y+1)! / (y+2)

Hope this helps!!

q = x!*y!/x!+y! = Integer?

Since (1) says x-y=1

x, y are consecutive integers

3!+4! can be written as 3!(4+1)
Similarly x!+y! = y!(x+1)

=> q = x!*y!/y!(x+1) = x!/x+1

x!/x+1 is an integer only if x is an odd integer greater than 3

(1) x-y = 1 (odd) so we know that out of x and y 1 is odd and the other even. This is not enough.

(2) The info. in (2) alone is insufficient.

(1) and (2):

(2) tells us y is an even number but it is not 2 but an even number which is greater than 2 (it cannot be 0 as one can't dig a well in 0 hours). thus x-y =1 is not satisfied by x=2 and y=3.

Thus, y is is E and x is O (greater than 3). Thus, q = x!/x+1 we can conclude is an Integer based on (1) and (2). Ans. (C).

Hi,

Can you please explain the following part a bit more clearly -

"Taking y! as common in both numerator and denominator"

q = y!*(y+1)! / y!*(y+1 +1)

=> q = (y+1)! / (y+2)

Simply reduce \(\frac{y!*(y+1)!}{y!*(y+1+1)}\) by y! to get \(\frac{(y+1)!}{(y+1+1)}=\frac{(y+1)!}{(y+2)}\).

Is there a less time consuming solution?

B = \(\frac{1}{x!}\)
C = \(\frac{1}{y!}\)
Together = \(\frac{1}{q}\)
=> q = \(\frac{x!y!}{x!+y!}\)
q = Integer ?

1) x - y = 1
x = 3, y = 2
q is not an integer
x = 4. y = 3
q = \(\frac{4*3!*3!}{5*3!}\) = not integer
x = 5, y = 4 => q = 20 (Integer)
Insufficient.

2) y is an even number other than 2
Insufficient. We know nothing about x

1+2)
x - y = 1
and y = even except 2
=> for all even values of y, q = integer.
x = 5, y = 4 => \(\frac{5*4!*4!}{6*4!}\) = 20
x = 7, y = 6 => \(\frac{7*6!*6!}{8*6!}\) = 630
Sufficient. Answer is C

I have a different perspective, but someone let me know if I am off base. I tried to be more logical, than math based.
I set x=a+2 and y=a+3 and plugged for 1/x!+1/y!=1/q... 1/5!+1/4!--> 1/120+5/120 =6/120 = 20

I saw it this way. Any number you will get for \frac{1}{x!} will be odd and added to the \frac{1}{y!}. Any total factorial will be divisible by that number because it has factors built into the total.

To the above - 5! = 1*2*3*4*5 = 120 --- This must be divisible by 6 because 2*3 in the equation equals 6.

By giving that x! is a prime number, and we now our 1/x!+1/y! will always give us an even number. Any denominator will have factor that multiply to equal the numerator.

1/7!+1/6! -->1/5040+7/5040 = 8/5040 Because 2*4 is a part of 7!, we know it will result in an integer.

It makes sense in my madness...

The stem tells us that x+y=q.
--Essentially I just need to know if x and y are integers or if they add together to form an integer.

(1) tells us that x-y=1.
--This doesn't provide the answer I need, it just says that x and y are 1 apart.

(2) basically just tells us that y is an integer.
--This doesn't tell me anything about x or q so it doesn't provide the info I need either.

(1 & 2)
I used a combination method to solve using the info from the stem and statement one to do the following:
x-y=1
-x+y=q
-2y=1-q
2y+1=q

Now, using info from statement 2, we know that y is an integer and doubling an integer and adding 1 to it will give us an integer.

Hope this makes sense. Please let me know if you spot any errors! Thanks,

Can you please give a easy way to do this
VeritasKarishma
GMATinsight

Shouldn't the answer be E,
Y can be 0, because 0! = 1, and while number of hours cannot be 0, y is not number of hours.
Also, 0 is a non prime even number without a doubt.
if Y = 0;
Q = 1/2.