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08 Jul 2024, 08:50
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
57% (01:51) correct
43%
(01:41)
wrong
based on 218
sessions
History
Date
Time
Result
Not Attempted Yet
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
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??
Can anyone help??
08 Jul 2024, 09:54
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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
\((\sqrt{15!} + \sqrt{16!})^2\)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??
Can anyone help??
\((\sqrt{15!})^2 + (\sqrt{16!})^2+2(\sqrt{15!} * \sqrt{16!})\)
\(15! + 16!+2(15! * \sqrt{16})\)
\(15! + 16!+2(15! * 4)\)
\(15!(1 + 16+8)=15!*25=1*2*...*5*...*10*....*15*25\)
So, it contains 5^5 in it.
Answer = 5
08 Jul 2024, 12:47
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So, you can write \((\sqrt{15!} + \sqrt{16!})^2\) as \((\sqrt{15!} + \sqrt{15!*16})^2\)
The square root of 16 is 4 so we can write this as \((\sqrt{15!} + 4\sqrt{15!})^2\) which is \((5\sqrt{15!})^2\)
Once we open the brackets we get 25*15!. The question asks what the greatest possible value of n is such that 25*15! is divisible by 5^n.
So basically, since 5 is a prime number how many times can 5 multiply by itself, so that it is divisible by 25*15!.
Well let us just break down 25*15!
25 is 5*5. 15! as we know is the multiple of all integers from 15 to 1. How many of those integers contain a 5? Well, 5, 10 (5*2) and 15 (5*3). So basically we could rewrite 25*15! as 5*5*5*5*5*1*2*3*4*6*7*8...... (no need to write these numbers out since its a waste of time and we only care about the 5s)
We can see that there are five 5s, in other words 5^5. Therefore, the greatest possible value of n is 5.
The square root of 16 is 4 so we can write this as \((\sqrt{15!} + 4\sqrt{15!})^2\) which is \((5\sqrt{15!})^2\)
Once we open the brackets we get 25*15!. The question asks what the greatest possible value of n is such that 25*15! is divisible by 5^n.
So basically, since 5 is a prime number how many times can 5 multiply by itself, so that it is divisible by 25*15!.
Well let us just break down 25*15!
25 is 5*5. 15! as we know is the multiple of all integers from 15 to 1. How many of those integers contain a 5? Well, 5, 10 (5*2) and 15 (5*3). So basically we could rewrite 25*15! as 5*5*5*5*5*1*2*3*4*6*7*8...... (no need to write these numbers out since its a waste of time and we only care about the 5s)
We can see that there are five 5s, in other words 5^5. Therefore, the greatest possible value of n is 5.
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
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??
Can anyone help??
