BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
Vote for Target Test Prep, Newsweek Readers’ Choice Awards 2026
NEWSWEEK READERS’ CHOICE 2026

BIG NEWS! Target Test Prep has been nominated, and they’d love your vote!

TTP has worked incredibly hard to build the best test prep experience possible, and winning Newsweek’s 2026 Readers’ Choice Award for Best Test Prep would mean a lot to them. If TTP has helped you, they’d be incredibly grateful for your vote. You can vote once each day through September 9.

Vote for TTP
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Number and Integers

Expert replies
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]
Join the discussion
Source: — Data Sufficiency |

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...
Join the discussion

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
Join the discussion

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)
Join the discussion

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!
Join the discussion