NEW HERE? LEARN MORE
closeclose
Last visit was: 13 May 2024, 01:57 It is currently 13 May 2024, 01:57
Decision Tracker
My Rewards
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

The value of ((8!) + (9!))^2 is an integer. What is the greatest in

SORT BY:
Date
Kudos
Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 93210
Own Kudos [?]: 623317 [84]
Given Kudos: 81851
Send PM
CAT Tests Forum Quiz Mode
The value of ((8!) + (9!))^2 is an integer. What is the greatest in [#permalink] New post   17 Aug 2023, 03:00
5
Kudos
79
Bookmarks
Expert Reply
00:00
A
B
C
D
E

Difficulty:

75% (hard)

Question Stats:

62% (02:15) correct 38% (02:28) wrong based on 444 sessions
Hide Show timer Statistics
The value of \((\sqrt{8!} + \sqrt{9!})^2\) is an integer. What is the greatest integer n such that \(2^n\) is a factor of \((\sqrt{8!} + \sqrt{9!})^2\)?

A. 3
B. 6
C. 8
D. 11
E. 14­


Show Spoiler
ShowHide Answer
Official Answer
D
Most Helpful Reply
avatar
G
DG1989
Manager
Manager
Joined: 16 Feb 2023
Posts: 175
Own Kudos [?]: 159 [23]
Given Kudos: 7
Location: India
Concentration: Finance, Technology
Schools: Kellogg '26
GPA: 4
Send PM
Re: The value of ((8!) + (9!))^2 is an integer. What is the greatest in [#permalink] New post   17 Aug 2023, 04:16
13
Kudos
10
Bookmarks
Bunuel wrote:
The value of \((\sqrt{8!} + \sqrt{9!})^2\) is an integer. What is the greatest integer n such that \(2^n\) is a factor of \((\sqrt{8!} + \sqrt{9!})^2\)?

A. 3
B. 6
C. 8
D. 11
E. 14


\((\sqrt{8!} + \sqrt{9!})^2\) = (8! + 9! + 2 * \(\sqrt{8! * 9!}\))
= 8! + 9! + 2 * \(\sqrt{8! * 8! * 9}\)
= 8! + 9! + 2 * 3 * \(\sqrt{8! * 8!}\)
= 8! + 9! + 2 * 3 * 8!
= 8! * (1 + 9 + 6)
= 16 * 8!
= \(2^4\) * 8!

Now maximum power of 2 in 8! = 7
Hence total power of 2 in the given expression = \(2^4 * 2^7\)
= \(2^{11}\)

Thus for n = 11, \(2^n\) is a factor of \((\sqrt{8!} + \sqrt{9!})^2\)

Option D is the correct answer
General Discussion
User avatar
L
gmatophobia
Quant Chat Moderator
Joined: 22 Dec 2016
Posts: 3122
Own Kudos [?]: 4344 [3]
Given Kudos: 1853
Location: India
Concentration: Strategy, Leadership
Send PM
CAT Tests Forum Quiz Mode
Re: The value of ((8!) + (9!))^2 is an integer. What is the greatest in [#permalink] New post   17 Aug 2023, 03:46
2
Kudos
1
Bookmarks
Bunuel wrote:
The value of \((\sqrt{8!} + \sqrt{9!})^2\) is an integer. What is the greatest integer n such that \(2^n\) is a factor of \((\sqrt{8!} + \sqrt{9!})^2\)?

A. 3
B. 6
C. 8
D. 11
E. 14


\(x = 8! \)

\((\sqrt{8!} + \sqrt{9*8!})^2\)

\((\sqrt{x} + \sqrt{9x})^2\)

\(x + 9x + 2\sqrt{x}*\sqrt{9x} \)

\(10x + 2\sqrt{x * 9x}\)

\(10x + 2\sqrt{9x^2}\)

\(10x + 6x\)

\(16x\)

\(16*8!\)

Power or 2 in 16 = 4 (because \(2^4 = 16\))

Power of 2 in 8! = 7
Show Spoiler
8/2 = 4
4/2 = 2
2/2 = 1

4 + 2 + 1 = 7

Power of 2 in (\(16*8!\)) = \(2^4 * 2^7 = 2^{11}\)

Option D
User avatar
S
Paras96
Director
Director
Joined: 11 Sep 2022
Posts: 500
Own Kudos [?]: 156 [3]
Given Kudos: 2
Location: India
Paras: Bhawsar
GMAT 1: 590 Q47 V24
GMAT 2: 580 Q49 V21
GMAT 3: 700 Q49 V35
GPA: 3.2
WE:Project Management (Other)
Send PM
Re: The value of ((8!) + (9!))^2 is an integer. What is the greatest in [#permalink] New post   17 Aug 2023, 05:18
3
Kudos
N=(√8!+√9!)2
N=(√8!+√9*8!)2
N= (√8!+3√8!)2
N=(4√8!)2
N=16*8!

8/2=4/2=2/2=1

N= (2^4)*(2^7)*p=(2^11)*p

Therefore , n =11

Hence D
User avatar
P
JeffTargetTestPrep
Target Test Prep Representative
Joined: 04 Mar 2011
Status:Head GMAT Instructor
Affiliations: Target Test Prep
Posts: 3042
Own Kudos [?]: 6340 [4]
Given Kudos: 1646
Send PM
Forum Quiz Mode
Re: The value of ((8!) + (9!))^2 is an integer. What is the greatest in [#permalink] New post   20 Aug 2023, 17:20
2
Kudos
2
Bookmarks
Expert Reply
Bunuel wrote:
The value of \((\sqrt{8!} + \sqrt{9!})^2\) is an integer. What is the greatest integer n such that \(2^n\) is a factor of \((\sqrt{8!} + \sqrt{9!})^2\)?

A. 3
B. 6
C. 8
D. 11
E. 14


(√8! + √9!)^2 = 8! + 2√(8!×9!) + 9! =

8! + 2√(8!×8!×9) + 9! =

8! + 2×8!×3 + 9! =

8!(1 + 6 + 9) =

8!×16 =

8!×2^4

We need to find the greatest integer n such that (8!×2^4)/2^n = integer. The expression above is an integer if n is not greater than the total number of 2s in the prime factored form of the numerator.

We can quickly determine the total number of 2s in the prime factored form of 8! with the following technique:

8/2 = 4; 4/2 = 2; 2/2 = 1

The total number of 2s is 4 + 2 + 1 = 7, so we have:

n ≤ 7 + 4 [We add 4 because there are 4 additional 2s in 2^4.]

n ≤ 11

We see that the maximum value of integer n is 11.

Answer: D
User avatar
B
sanyashah
Intern
Intern
Joined: 08 Sep 2023
Posts: 28
Own Kudos [?]: 4 [0]
Given Kudos: 205
Location: India
Send PM
Re: The value of ((8!) + (9!))^2 is an integer. What is the greatest in [#permalink] New post   01 Jan 2024, 10:10
Can someone explain the last step please?

This one:
Power of 2 in 8!
8/2 + 8/4+ 8/8 = 4+2+1 =7
Power of 16 = 4
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 93210
Own Kudos [?]: 623317 [0]
Given Kudos: 81851
Send PM
CAT Tests Forum Quiz Mode
Re: The value of ((8!) + (9!))^2 is an integer. What is the greatest in [#permalink] New post   01 Jan 2024, 10:40
Expert Reply
sanyashah wrote:
Can someone explain the last step please?

This one:
Power of 2 in 8!
8/2 + 8/4+ 8/8 = 4+2+1 =7
Power of 16 = 4


This is explained here Everything about Factorials on the GMAT.
User avatar
B
ashdank94
Intern
Intern
Joined: 20 Feb 2023
Posts: 31
Own Kudos [?]: 7 [0]
Given Kudos: 102
Send PM
Re: The value of ((8!) + (9!))^2 is an integer. What is the greatest in [#permalink] New post   14 Jan 2024, 16:46
Hello, could someone help me understand why the 16 has 4 2's? When I divide it by 2 it gives me 8 so I add it to the 7 2's I get in 8!
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 93210
Own Kudos [?]: 623317 [0]
Given Kudos: 81851
Send PM
CAT Tests Forum Quiz Mode
Re: The value of ((8!) + (9!))^2 is an integer. What is the greatest in [#permalink] New post   15 Jan 2024, 00:21
Expert Reply
ashdank94 wrote:
Hello, could someone help me understand why the 16 has 4 2's? When I divide it by 2 it gives me 8 so I add it to the 7 2's I get in 8!


When finding the highest integer power of a prime number p, in the n!, the factorial of a positive integer n, we calculate the following:

    \(\frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3}+...+\frac{n}{p^k}\), where \(k\) must be chosen such that \(p^k\leq{n}\)

Example:
What is the power of 2 in 25!?

    \(\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22\).

Observe that we take only the quotient of the division above, that is for example, 25/8 = 3.

However, in the problem at hand we have 16, not the factorial of 16, so since 16 = 2^4, then the highest integer power of 2 in 16 is 4. We need not use the formula above. If it were 16! instead, then the highest integer power of 2 in 16! would have been 16/2 + 16/4 + 16/8 + 16/16 = 8 + 4 + 2 + 1 = 15.

Check foe more here: Everything about Factorials on the GMAT.

Hope it's clear.
User avatar
B
Jake7Wimmer
Intern
Intern
Joined: 25 Apr 2023
Posts: 23
Own Kudos [?]: 5 [0]
Given Kudos: 5
GMAT 1: 650 Q48 V32
GMAT 2: 650 Q47 V32
Send PM
Re: The value of ((8!) + (9!))^2 is an integer. What is the greatest in [#permalink] New post   10 Mar 2024, 12:36
 
JeffTargetTestPrep wrote:
Bunuel wrote:
The value of \((\sqrt{8!} + \sqrt{9!})^2\) is an integer. What is the greatest integer n such that \(2^n\) is a factor of \((\sqrt{8!} + \sqrt{9!})^2\)?

A. 3
B. 6
C. 8
D. 11
E. 14

(√8! + √9!)^2 = 8! + 2√(8!×9!) + 9! =

8! + 2√(8!×8!×9) + 9! =

8! + 2×8!×3 + 9! =

8!(1 + 6 + 9) =

8!×16 =

8!×2^4

We need to find the greatest integer n such that (8!×2^4)/2^n = integer. The expression above is an integer if n is not greater than the total number of 2s in the prime factored form of the numerator.

We can quickly determine the total number of 2s in the prime factored form of 8! with the following technique:

8/2 = 4; 4/2 = 2; 2/2 = 1

The total number of 2s is 4 + 2 + 1 = 7, so we have:

n ≤ 7 + 4 [We add 4 because there are 4 additional 2s in 2^4.]

n ≤ 11

We see that the maximum value of integer n is 11.

Answer: D

­JeffTargetTestPrep

This step right here I missed on practice test. We obviously do that a lot with exponents and I haven't really seen this to much with factorial related questions. How would we know to pull out the common factors? Or is this just something that is a given we should know what to do. 

8! + 2×8!×3 + 9! =

8!(1 + 6 + 9) =
User avatar
L
KarishmaB
Tutor
Joined: 16 Oct 2010
Posts: 14891
Own Kudos [?]: 65253 [1]
Given Kudos: 431
Location: Pune, India
Send PM
Forum Quiz Mode
Re: The value of ((8!) + (9!))^2 is an integer. What is the greatest in [#permalink] New post   08 Apr 2024, 07:57
1
Kudos
Expert Reply
 
Bunuel wrote:
The value of \((\sqrt{8!} + \sqrt{9!})^2\) is an integer. What is the greatest integer n such that \(2^n\) is a factor of \((\sqrt{8!} + \sqrt{9!})^2\)?

A. 3
B. 6
C. 8
D. 11
E. 14

­\((\sqrt{8!} + \sqrt{9!})^2\)

\(=(\sqrt{8!} + \sqrt{9*8!})^2\)

\(=(\sqrt{8!} + 3*\sqrt{8!})^2\)

\(=(4 * \sqrt{8!})^2\)

\(= 16*8!\)

\(= 2^4 * 8!\)

We know how to find the highest power in a factorial. The logic is discussed in the link of the post given below.
8/2 + 4/2 + 2/2 = 7

Hence we get have 4 2s from 2^4 and 7 2s from 8! giving us a total of 11 2s

Answer (D)

How to find maximum power in a factortial:
https://anaprep.com/number-properties-h ... actorials/
 ­
User avatar
B
DanTheGMATMan
Tutor
Joined: 02 Oct 2015
Posts: 90
Own Kudos [?]: 23 [0]
Given Kudos: 4
Send PM
The value of ((8!) + (9!))^2 is an integer. What is the greatest in [#permalink] New post   25 Apr 2024, 13:04
Expert Reply
Try following this here:
­
GMAT Club Bot
gmatclubot
The value of ((8!) + (9!))^2 is an integer. What is the greatest in [#permalink]
25 Apr 2024, 13:04
Moderators:
Bunuel
Math Expert
93206 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3136 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions