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Bill can dig a well in x! hours. Carlos can dig the same [#permalink]

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22 Jul 2010, 00:02

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Bill can dig a well in x! hours. Carlos can dig the same well in y! hours. If q is the number of hours that it takes Bill and Carlos to dig the well together, working at their respective rates, is q an integer?

Thanks Gurpreet! But can you please explain the below steps a bit more. Lost track and couldnt follow u. Statement 1: x -y=1 => x = y+1 => x! = (y+1)y! and the next step as well!

Thanks Gurpreet! But can you please explain the below steps a bit more. Lost track and couldnt follow u. Statement 1: x -y=1 => x = y+1 => x! = (y+1)y! and the next step as well!

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

Thanks Gurpreet! But can you please explain the below steps a bit more. Lost track and couldnt follow u. Statement 1: x -y=1 => x = y+1 => x! = (y+1)y! and the next step as well!

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

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

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

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).

Re: Bill can dig a well in x! hours. Carlos can dig the same [#permalink]

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02 Oct 2014, 20:06

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Re: Bill can dig a well in x! hours. Carlos can dig the same [#permalink]

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04 Jan 2016, 05:52

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Bill can dig a well in x! hours. Carlos can dig the same [#permalink]

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21 Jul 2017, 04:09

rohitgoel15 wrote:

Bill can dig a well in x! hours. Carlos can dig the same well in y! hours. If q is the number of hours that it takes Bill and Carlos to dig the well together, working at their respective rates, is q an integer?

(1) x - y = 1 (2) y is a nonprime even number.

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
_________________

if y = 2 q = 3!/4 = 3/2 not an integer if y = 4 Q= 5!/6 = 20 = integer

Thus not sufficient.

Considering both the statement y cannot be 2, for all even values of y except 2, q is an integer.

Thus C

Is there a less time consuming solution?

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.