Number and Integers

This topic has expert replies
User avatar
Master | Next Rank: 500 Posts
Posts: 123
Joined: Tue Nov 23, 2010 7:18 pm
Location: Mumbai, India
Thanked: 5 times
Followed by:4 members

Number and Integers

by anirudhbhalotia » Wed Dec 15, 2010 9:11 pm
P.S. - Question source, from GMAT Hacks!


If x is a positive integer, what is the result when (x + 1)! is divided by (x - 1)! ?
(1) (x - 1)! = 720
(2) x(x + 1) = 56

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.


Answer with explanation -

[spoiler]Answer: D
While factorials can be time-consuming to calculate, we don't need to find the exact answer on a Data Sufficiency question. In this case, it is sufficient to know that we could solve. To do so, all we would need is the value of x.

Statement (1) is sufficient. Given one variable, as we are here, we can determine which value of x - 1 has a factorial equivalent to 720. (It turns out that if x = 7, x - 1 = 6, and 6! = 720.)

Statement (2) is also sufficient. We can determine from the statement that x = 7. (Remember, x must be positive.) That's enough to answer the question. Choice (D) is sufficient.

One more note. On more advanced questions, it might be handy to recognize that (x + 1)! = (x + 1)(x)(x - 1)!, so when it is divided by (x - 1)!, the result is (x + 1)(x)--statement (2). If you needed to solve for (x + 1)! divided by (x - 1)!, that would be a simple way of reaching a solution.[/spoiler]
Source: — Data Sufficiency |

User avatar
Master | Next Rank: 500 Posts
Posts: 123
Joined: Tue Nov 23, 2010 7:18 pm
Location: Mumbai, India
Thanked: 5 times
Followed by:4 members

by anirudhbhalotia » Wed Dec 15, 2010 9:18 pm
I chose B as the answer.

1. (x - 1)! = 720

If the no. is too big...say 10230 or something...how do we find the value of X...by doing manual calculation!


2. x(x + 1) = 56
This looks simple than the above and much more easy to solve. Thats why I chose B as the answer.



Am I missing the point here...even though first statement looks difficult to solve in-case a no. is big...its still doable...or should I not assume and just go with whats in context at present?

This is a constant confusion for me...

Master | Next Rank: 500 Posts
Posts: 161
Joined: Mon Apr 05, 2010 9:06 am
Location: Mumbai
Thanked: 37 times

by 4GMAT_Mumbai » Wed Dec 15, 2010 11:08 pm
Hi Anirudh,

If you know that (x - 1) ! = 720; then essentially one can find the value of X uniquely.

The question is asking for (x + 1)! / (x - 1)! which boils down to (x+1) * x

If you can find out x; then you can figure out x*(x+1) also. Hence, sufficient. There is no necessity to actually do the calculations.

For example, even if the question had been (x - 1)! = 362,880; the rationale would have just been the same and no different.

Catch: Just make sure that the answer will be a unique one.

If the question is "What is x^2?" and the 1st statement says

(x-20) * (x -30) = 0;

then, I will get two different values for x^2.

Hence, the two questions you should ask yourself when calculations get too complicated are:

1) Will this lead me to answer the question (eventually)?

2) Will the answer be an unique one?

If so, sufficient ! I hope this helps with your constant confusion ;-)
Naveenan Ramachandran
4GMAT, Dadar(W) & Ghatkopar(W), Mumbai

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 1179
Joined: Sun Apr 11, 2010 9:07 pm
Location: Milpitas, CA
Thanked: 447 times
Followed by:88 members

by Rahul@gurome » Wed Dec 15, 2010 11:12 pm
anirudhbhalotia wrote:I chose B as the answer.

1. (x - 1)! = 720

If the no. is too big...say 10230 or something...how do we find the value of X...by doing manual calculation!


2. x(x + 1) = 56
This looks simple than the above and much more easy to solve. Thats why I chose B as the answer.



Am I missing the point here...even though first statement looks difficult to solve in-case a no. is big...its still doable...or should I not assume and just go with whats in context at present?

This is a constant confusion for me...
I think you mean 40320 instead of 10230.
If (x-1)! = 40320, the following method may be helpful.
Note that 40320 is 4032*2*5.
So the (x-1)! has only one 5 in it.
So 5 <= (x-1) <10.
So (x-1) can be 5, 6, 7, 8, 9.
Since 7 is a prime, check if 7 divides 4032.
You get 40320 = 7*576*2*5.
So 7<= (x-1) < 10.
Or (x-1) = 7, 8 or 9.
If (x-1)! is 8! it should be divisible by 2*4*6*8.
Or It will have 2, (1+2+1+3) = 7 times.
Or 40320 should be divisible by 2^7 = 128 which it is.
If (x-1)! is 9!, it will be divisible by 3*6*9.
Or it should have 3, (1+1+2) = 4 times.
Or 40320 should be divisible by 3^4 which it is not.
So (x-1)! cannot be 9!.
Or (x-1)! = 8!.
Or x = 9.
So even if we have a large value, the problem is doable from statement (1) alone.
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)

User avatar
Master | Next Rank: 500 Posts
Posts: 123
Joined: Tue Nov 23, 2010 7:18 pm
Location: Mumbai, India
Thanked: 5 times
Followed by:4 members

by anirudhbhalotia » Wed Dec 15, 2010 11:21 pm
Rahul@gurome wrote:
anirudhbhalotia wrote:I chose B as the answer.

1. (x - 1)! = 720

If the no. is too big...say 10230 or something...how do we find the value of X...by doing manual calculation!


2. x(x + 1) = 56
This looks simple than the above and much more easy to solve. Thats why I chose B as the answer.



Am I missing the point here...even though first statement looks difficult to solve in-case a no. is big...its still doable...or should I not assume and just go with whats in context at present?

This is a constant confusion for me...
I think you mean 40320 instead of 10230.
If (x-1)! = 40320, the following method may be helpful.
Note that 40320 is 4032*2*5.
So the (x-1)! has only one 5 in it.
So 5 <= (x-1) <10.
So (x-1) can be 5, 6, 7, 8, 9.
Since 7 is a prime, check if 7 divides 4032.
You get 40320 = 7*576*2*5.
So 7<= (x-1) < 10.
Or (x-1) = 7, 8 or 9.
If (x-1)! is 8! it should be divisible by 2*4*6*8.
Or It will have 2, (1+2+1+3) = 7 times.
Or 40320 should be divisible by 2^7 = 128 which it is.
If (x-1)! is 9!, it will be divisible by 3*6*9.
Or it should have 3, (1+1+2) = 4 times.
Or 40320 should be divisible by 3^4 which it is not.
So (x-1)! cannot be 9!.
Or (x-1)! = 8!.
Or x = 9.
So even if we have a large value, the problem is doable from statement (1) alone.

No...I didn't intend for 40320 specifically. I just intended for any big no. and how to find x from that...But I see what you mean!