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Bunuel
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The value of ((8!) + (9!))^2 is an integer. What is the greatest in 23 Jun 2025, 23:39
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Sushi_545
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

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)
Show more

Have no idea what you are doing there...
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Sushi_545
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The value of ((8!) + (9!))^2 is an integer. What is the greatest in 23 Jun 2025, 23:48
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Bunuel
Sushi_545
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

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)

Have no idea what you are doing there...
Show more


Hi bunuel,

Please find my attached image its hard to write here.
Attachments

20250624_121729.jpg
20250624_121729.jpg [ 4.62 MiB | Viewed 592 times ]

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APram
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The value of ((8!) + (9!))^2 is an integer. What is the greatest in 24 Jun 2025, 00:00
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Hi
In the equation that you are dealing with, you are not taking into consideration that factors of 2 will be there in addition of \(\sqrt{x} + \sqrt{y}\)

Hence it's an insufficient consideration


Sushi_545
Bunuel
Sushi_545
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

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)

Have no idea what you are doing there...


Hi bunuel,

Please find my attached image its hard to write here.
Show more
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GmatKnightTutor
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The value of ((8!) + (9!))^2 is an integer. What is the greatest in 24 Jun 2025, 00:04
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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


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

Replacing 8! with "a" and noticing 9! is basically 9(8!) or 9a can help a bit.

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

When we open this, we get:

a + 9a + 2\(\sqrt{a} \sqrt{9a}\)

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

a + 9a + 2(3a)

16a

Replace a with 8!

16(8!)

How many factors of 2 can go into this?

16 is equal to 2^4 so we have 4 there - and 8! has 7

4 + 7 = 11

(D) is your answer.
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The value of ((8!) + (9!))^2 is an integer. What is the greatest in 24 Jun 2025, 00:06
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Bunuel
Sushi_545
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

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)

Have no idea what you are doing there...


Hi bunuel,

Please find my attached image its hard to write here.
Show more

We are told that "The value of \((\sqrt{8!} + \sqrt{9!})^2\) is an integer", so \((\sqrt{x} + \sqrt{y})^2\) must be an integer, and could be even, contributing more powers of 2. So you cannot conclude that 7 is the highest power from that approach.
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The value of ((8!) + (9!))^2 is an integer. What is the greatest in 14 Jul 2025, 18:59
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Hi, can someone explain where the (1+6+9) comes from?
JeffTargetTestPrep
Bunuel
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
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Bunuel
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The value of ((8!) + (9!))^2 is an integer. What is the greatest in 14 Jul 2025, 22:50
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noodlebob
Hi, can someone explain where the (1+6+9) comes from?
JeffTargetTestPrep
Bunuel
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
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8! + 2*8!*3 + 9! =

= 8! + 8!*6 + 8!*9 =

= 8!(1 + 6 + 9)

Hope it helps.
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