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chetan2u

Aarushi225
What is the greatest possible value of n such that \((\sqrt{15!} + \sqrt{16!})^2\) is divisible by 5^n?

A. 0
B. 1
C. 3
D. 4
E. 5

The correct answer is 5 but i am getting 3 as my answer. 

Can anyone help??­
­\((\sqrt{15!} + \sqrt{16!})^2\)
\((\sqrt{15!})^2 + (\sqrt{16!})^2+2(\sqrt{15!} * \sqrt{16!})\)
\(15! + 16!^2+2(15! * \sqrt{16})\)
\(15! + 16!^2+2(15! * 4)\)
\(15!(1 + 16+8)=15!*25=1*2*...*5*...*10*....*15*25\)
So, it contains 5^5 in it.

Answer = 5
­Why has the (\sqrt {16!})^2 stayed 16!^2?
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chetan2u

Aarushi225
What is the greatest possible value of n such that \((\sqrt{15!} + \sqrt{16!})^2\) is divisible by 5^n?

A. 0
B. 1
C. 3
D. 4
E. 5

The correct answer is 5 but i am getting 3 as my answer. 

Can anyone help??­
­\((\sqrt{15!} + \sqrt{16!})^2\)
\((\sqrt{15!})^2 + (\sqrt{16!})^2+2(\sqrt{15!} * \sqrt{16!})\)
\(15! + 16!^2+2(15! * \sqrt{16})\)
\(15! + 16!^2+2(15! * 4)\)
\(15!(1 + 16+8)=15!*25=1*2*...*5*...*10*....*15*25\)
So, it contains 5^5 in it.

Answer = 5
­Why has the (\sqrt {16!})^2 stayed 16!^2?
­Hi
thanks.
That's a typing error due to cut and paste. In the final step, I did take it as 16! alone.
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does anyone have an answer that doesnt skip the most vital steps?
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onlymalapink
does anyone have an answer that doesnt skip the most vital steps?
https://gmatclub.com/forum/what-is-the- ... l#p3420319 - this post above has all the steps mentioned leading to the solution.
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Where has the factorial sign of 16 in 3rd line
chetan2u
Aarushi225
What is the greatest possible value of n such that \((\sqrt{15!} + \sqrt{16!})^2\) is divisible by 5^n?

A. 0
B. 1
C. 3
D. 4
E. 5

The correct answer is 5 but i am getting 3 as my answer.

Can anyone help??­
­\((\sqrt{15!} + \sqrt{16!})^2\)
\((\sqrt{15!})^2 + (\sqrt{16!})^2+2(\sqrt{15!} * \sqrt{16!})\)
\(15! + 16!+2(15! * \sqrt{16})\)
\(15! + 16!+2(15! * 4)\)
\(15!(1 + 16+8)=15!*25=1*2*...*5*...*10*....*15*25\)
So, it contains 5^5 in it.

Answer = 5­
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shubhim20
Where has the factorial sign of 16 in 3rd line
chetan2u
Aarushi225
What is the greatest possible value of n such that \((\sqrt{15!} + \sqrt{16!})^2\) is divisible by 5^n?

A. 0
B. 1
C. 3
D. 4
E. 5

The correct answer is 5 but i am getting 3 as my answer.

Can anyone help??­
­\((\sqrt{15!} + \sqrt{16!})^2\)
\((\sqrt{15!})^2 + (\sqrt{16!})^2+2(\sqrt{15!} * \sqrt{16!})\)
\(15! + 16!+2(15! * \sqrt{16})\)
\(15! + 16!+2(15! * 4)\)
\(15!(1 + 16+8)=15!*25=1*2*...*5*...*10*....*15*25\)
So, it contains 5^5 in it.

Answer = 5­

\((\sqrt{15!})^2 + (\sqrt{16!})^2+2(\sqrt{15!} * \sqrt{16!})\)

\((\sqrt{15!})^2 + (\sqrt{16!})^2+2(\sqrt{15!} * \sqrt{15!*16})\)

\((\sqrt{15!})^2 + (\sqrt{16!})^2+2(\sqrt{15!*15!*16})\)

\(15! + 16!+2(15! * \sqrt{16})\)

Hope it's clear.
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[square root(!15) + square root(!16)]^2
=!15+!16+2*square root(!15*!16)
=!15+16*!15+2*square root(!15*!15*16)
=!15+16*!15+2*4*!15
=!15(1+16+8)
=!15*25
power of 5 in the above expression=3+2=5
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