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

Question

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 https://gmatclub.com/forum/download/file.php?id=81452 GMAT-Club-Forum-yil0qgvt.png

DG1989 · Accepted Answer

(\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

Paras96 · Answer

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

gmatophobia · Answer

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
 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

JeffTargetTestPrep · Answer

(√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

n ≤ 11

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

Answer: D

mbaaspirantss07 · Answer

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

Bunuel · Answer

This is explained here  Everything about Factorials on the GMAT .

ashdank94 · Answer

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!

Bunuel · Answer

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.

KarishmaB · Answer

(\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/

DanTheGMATMan · Answer

Try following this here:
 https://www.youtube.com/watch?v=U7M8ixleGJA

andrewjohn8 · Answer

Thought I'd add my solution since I believe I didn't see it anywhere.

I factored out the Sqrt(8!) first.

(sqrt(8!(1 + sqrt(9))^2

which reduces to

((sqrt(8!)*(4))^2

8!*16

Then to do the same method as others have shown to get that there are 7 factors of 2 in 8! plus the 4 factors of 2 in 16 means there are at most 11 factors of 2.

Danou · Answer

I am having trouble understanding why there 7 "power of 2" in 8!. Could you further break it down please

KarishmaB · Answer

Discussed in detail here: https://anaprep.com/number-properties-h ... actorials/

Dbrunik · Answer

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

This line is quite confusing. specifically the part with the two factorials under the radical (2xy)

here is a better way to break it down.

hastyporpoise55 · Answer

Hi - would appreciate if anyone is able to clarify this for me ... i understand most of the logic in the above workings, except line 2 ... how does 2 (8!^1/2 * 9!^1/2) equate to what you then equated it to? thanks!

Bunuel · Answer

9! = 8! * 9, so:

\sqrt{8! * 9!} = \sqrt{8! * (8! * 9)}=3\sqrt{8! * 8!}=3 *8! .

lindsey123 · Answer

thank you bunuel, this is really helpful!

lavanya.18 · Answer

8!+9!+2 root 8!*9!
......root 8!*8!*9
.......2*8! * root9
.......2*8!*3
.........6*8!
8!+9!+6*8!
7*8!+9!
8!(7+9)
8!*16
8!*2^4

now 8! has 2^7

4+7=11 power of 2 needed------ANS!

Sushi_545 · Answer

Hi  Bunuel ,

what is wrong here?

8!= has 2 ^7 max power of 7
9! has 2^7 max power of 7

so (2^7/2(root(x)+root(y)))^2 translates to 2^7 (7 is the max power of 2)

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards