Last visit was: 11 Sep 2024, 05:03 It is currently 11 Sep 2024, 05:03
Decision Tracker
My Rewards
Unanswered
Active Topics
New posts
New comers' posts
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

If (n+2)!= n!(an^2+bn+c), then abc=?

SORT BY:
Date
Kudos
Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
L
MathRevolution
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 10131
Own Kudos [?]: 17262 [18]
Given Kudos: 4
GMAT 1: 760 Q51 V42
GPA: 3.82
Send PM
If (n+2)!= n!(an^2+bn+c), then abc=? [#permalink] If (n+2)!= n!(an^2+bn+c), then abc=?   12 Mar 2018, 01:04
1
Kudos
17
Bookmarks
Expert Reply
00:00
A
B
C
D
E

Difficulty:

25% (medium)

Question Stats:

74% (01:23) correct 26% (01:50) wrong based on 226 sessions
Hide Show DIFFICULTY AND TIMER STATISTICS
[GMAT math practice question]

If \((n+2)!= n!(an^2+bn+c),\) then \(abc=?\)

\(A. 2\)
\(B. 3\)
\(C. 4\)
\(D. 6\)
\(E. 8\)
ShowHide Answer
Official Answer
D
Most Helpful Reply
User avatar
L
BrentGMATPrepNow
GMAT Club Legend
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 6803
Own Kudos [?]: 31266 [4]
Given Kudos: 799
Location: Canada
Send PM
If (n+2)!= n!(an^2+bn+c), then abc=? [#permalink] If (n+2)!= n!(an^2+bn+c), then abc=?   Updated on: 26 Apr 2020, 13:50
1
Kudos
3
Bookmarks
Expert Reply
Top Contributor
MathRevolution wrote:
[GMAT math practice question]

If \((n+2)!= n!(an^2+bn+c),\) then \(abc=?\)

\(A. 2\)
\(B. 3\)
\(C. 4\)
\(D. 6\)
\(E. 8\)


First recognize that (n + 2)! = (n + 2)(n + 1)(n)(n - 1)(n - 2)(n - 3)....(3)(2)(1)
= (n + 2)(n + 1)(n!)
= (n!)(n + 2)(n + 1)

The question tells us that (n + 2)! = n!(an² + bn + c)
So, we can write: (n!)(n + 2)(n + 1) = n!(an² + bn + c)
Divide both sides by n! to get: (n + 2)(n + 1) = an² + bn + c
Use FOIL to expand left side: n² + 3n + 2 = an² + bn + c
In other words: 1n² + 3n + 2 = an² + bn + c

So, a = 1, b = 3 and c = 2
This means abc = (1)(3)(2) = 6

Answer: D

Cheers,
Brent

Originally posted by BrentGMATPrepNow on 14 Mar 2018, 09:08.
Last edited by BrentGMATPrepNow on 26 Apr 2020, 13:50, edited 1 time in total.
General Discussion
avatar
S
akshata19
Manager
Manager
Joined: 22 May 2017
Posts: 85
Own Kudos [?]: 113 [3]
Given Kudos: 105
Send PM
Re: If (n+2)!= n!(an^2+bn+c), then abc=? [#permalink] If (n+2)!= n!(an^2+bn+c), then abc=?   12 Mar 2018, 01:24
2
Kudos
1
Bookmarks
(n+2)(n+1)n!= n! (an2+bn+c)
(n+2)(n+1) = (an2+bn+c)
n2 + 3n +2 = (an2+bn+c)

abc = 1*3*2 =6

Posted from my mobile device
User avatar
L
MathRevolution
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 10131
Own Kudos [?]: 17262 [1]
Given Kudos: 4
GMAT 1: 760 Q51 V42
GPA: 3.82
Send PM
Re: If (n+2)!= n!(an^2+bn+c), then abc=? [#permalink] If (n+2)!= n!(an^2+bn+c), then abc=?   14 Mar 2018, 00:24
1
Bookmarks
Expert Reply
=>

\((n+2)! = (n+2)(n+1)n! = (n^2 + 3n + 2)n!\)
So,
\(a = 1, b = 3\), and \(c = 2\).
Thus, \(abc = 6.\)

Therefore, D is the answer.

Answer: D
User avatar
L
AkshdeepS
VP
VP
Joined: 13 Apr 2013
Status:It's near - I can see.
Posts: 1460
Own Kudos [?]: 1663 [1]
Given Kudos: 1001
Location: India
Concentration: International Business, Operations
GPA: 3.01
WE:Engineering (Real Estate)
Send PM
Reviews Badge
Re: If (n+2)!= n!(an^2+bn+c), then abc=? [#permalink] If (n+2)!= n!(an^2+bn+c), then abc=?   16 Mar 2018, 23:56
1
Kudos
MathRevolution wrote:
[GMAT math practice question]

If \((n+2)!= n!(an^2+bn+c),\) then \(abc=?\)

\(A. 2\)
\(B. 3\)
\(C. 4\)
\(D. 6\)
\(E. 8\)


It can be written as:

\(\frac{(n + 2) (n + 1) (n)!}{n!}\) \(=\) \((an^2+bn+c)\)

\((n + 2) (n + 1)\)= \((an^2+bn+c)\)

\(n^2 + 3n + 2 =\) \((an^2+bn+c)\)

Therefore, \(a = 1, b = 3, c = 2\)

(D)
User avatar
P
JS1290
Manager
Manager
Joined: 27 Dec 2016
Posts: 239
Own Kudos [?]: 245 [0]
Given Kudos: 1103
Send PM
Re: If (n+2)!= n!(an^2+bn+c), then abc=? [#permalink] If (n+2)!= n!(an^2+bn+c), then abc=?   09 Mar 2019, 19:59
Hi chetan2u,

I was wondering could you please explain where I am making a mistake in my solution? I am getting different answers.

Let's say n=1

6=1(1a+1b+c). If we were to pick a=b=1 and c=4, abc=6 BUT if we were to pick a=b=c=2, we get 8 instead.

Can we not pick smart numbers for this question specifically?
User avatar
L
chetan2u
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11513
Own Kudos [?]: 35932 [0]
Given Kudos: 333
Send PM
Reviews Badge
Re: If (n+2)!= n!(an^2+bn+c), then abc=? [#permalink] If (n+2)!= n!(an^2+bn+c), then abc=?   09 Mar 2019, 20:39
Expert Reply
JS1290 wrote:
Hi chetan2u,

I was wondering could you please explain where I am making a mistake in my solution? I am getting different answers.

Let's say n=1

6=1(1a+1b+c). If we were to pick a=b=1 and c=4, abc=6 BUT if we were to pick a=b=c=2, we get 8 instead.

Can we not pick smart numbers for this question specifically?


If you want to solve by plugging numbers here, you should look at for possible answer but then test these numbers in other values of n..
for example take n as 2 now..
\((2+2)!=2!(a2^2+2b+c).......24=2(4a+2b+c)..........12=4a+2b+c\)
the values of (a, b, c) when n=1 were (1, 1, 4) or (1,4,1) or (4,1,1) or (2,2,2) or (1,2,3) or (1,3,2) and so on..
Only (1,3,2) satisfies the equation when n=2.

So, the method of plugging would become a bit more cumbersome, although it will give you a correc answer after few steps
User avatar
bumpbot
Non-Human User
Joined: 09 Sep 2013
Posts: 34811
Own Kudos [?]: 876 [0]
Given Kudos: 0
Send PM
Re: If (n+2)!= n!(an^2+bn+c), then abc=? [#permalink] If (n+2)!= n!(an^2+bn+c), then abc=?   04 Mar 2024, 11:37
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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
GMAT Club Bot
gmatclubot
Re: If (n+2)!= n!(an^2+bn+c), then abc=? [#permalink]
04 Mar 2024, 11:37
Moderator:
Bunuel
Math Expert
95450 posts