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Joined: 20 Feb 2017
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What is the highest index of ‘2’ that can divide (10! + 11!
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28 Oct 2020, 01:14
28 Oct 2020, 01:14
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Question Stats:
41% (01:34) correct
58% (01:17) wrong based on 12 sessions
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What is the highest index of ‘2’ that can divide \((10! + 11! + 12!)\) without leaving any remainder?
(A) 8
(B) 9
(C) 12
(D) 16
(E) 25
(A) 8
(B) 9
(C) 12
(D) 16
(E) 25
Re: What is the highest index of ‘2’ that can divide (10! + 11!
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28 Oct 2020, 02:44
28 Oct 2020, 02:44
2
(10! + 11! + 12!)= 10!(1+11+132)=10!*144
in 10! index of 2 will be eight and 144 =16*9=\(2^4 *9\)
index of 2 will be 8+4=12
option C is correct
in 10! index of 2 will be eight and 144 =16*9=\(2^4 *9\)
index of 2 will be 8+4=12
option C is correct
Re: What is the highest index of ‘2’ that can divide (10! + 11!
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20 Feb 2021, 07:16
20 Feb 2021, 07:16
3
Okay, at first let's simplify little bit.
We have 10! + 11! + 12! = 10! + 11 * 10! + 12 * 11 * 10! = 10! (1 + 11 + 12 * 11) = 144 * 10!
Now, We will prime factor this simplified thing.
144 = 2 * 2 * 2 * 2 * 3 * 3
10! = 1 * 2 * 3 * 4* 5 * 6 * 7 * 8 * 9 * 10 = 1 * 2 * 3 * ( 2 * 2) * 5 * (2 * 3) * 7 * ( 2 * 2 * 2) * 9 * ( 2 * 5 )
So, this is very simple now, we have a total of 12 2's in 144 * 10!
That means this number will be divisible by 2^12 without leaving any reminder.
The answer is C.
We have 10! + 11! + 12! = 10! + 11 * 10! + 12 * 11 * 10! = 10! (1 + 11 + 12 * 11) = 144 * 10!
Now, We will prime factor this simplified thing.
144 = 2 * 2 * 2 * 2 * 3 * 3
10! = 1 * 2 * 3 * 4* 5 * 6 * 7 * 8 * 9 * 10 = 1 * 2 * 3 * ( 2 * 2) * 5 * (2 * 3) * 7 * ( 2 * 2 * 2) * 9 * ( 2 * 5 )
So, this is very simple now, we have a total of 12 2's in 144 * 10!
That means this number will be divisible by 2^12 without leaving any reminder.
The answer is C.
