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What is the greatest value of integer n such that 4^n is a factor of 2

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What is the greatest value of integer n such that 4^n is a factor of 2  [#permalink]

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New post 11 Jan 2017, 02:30
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A
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Difficulty:

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Question Stats:

51% (01:34) correct 49% (01:36) wrong based on 212 sessions

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Re: What is the greatest value of integer n such that 4^n is a factor of 2  [#permalink]

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New post 11 Jan 2017, 10:03
1
1
Bunuel wrote:
What is the greatest value of integer n such that 4^n is a factor of 26! ?

A. 6
B. 9
C. 10
D. 11
E. 12



no. of 2's in 26!
26/2=13
26/4=6
26/8=3
26/16=1

total 13+6+3+1=23
thus no of 4's(4^n)=23/2=11

Ans D
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Re: What is the greatest value of integer n such that 4^n is a factor of 2  [#permalink]

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New post 11 Jan 2017, 15:24
I listed all the factor divisible by 2 because odd number cannot be divisible by 4

2 4 6 8 10 12 14 16 18 20 22 24

4 8 12 16 20 24 it is all the multiple of 4 so we have => Here we can divide by 4 8 times 6 + 1 ( 16 can be divided 2 times ) + 1 (8 and 24 divided by 4 gives 2 and 6 which is again by 4 an another time)

2 6 10 14 18 22 26 it is all multiple of 2 but not 4 thus we need to make a pair to divide them by 4 (ex: 2*6=12 => Divided by 4) => Here we have 3 pairs possibles so we can divided 3 times

8 + 3 = 11 => D
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Re: What is the greatest value of integer n such that 4^n is a factor of 2  [#permalink]

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New post 11 Jan 2017, 18:12
LHC8717 wrote:
I listed all the factor divisible by 2 because odd number cannot be divisible by 4

2 4 6 8 10 12 14 16 18 20 22 24

4 8 12 16 20 24 it is all the multiple of 4 so we have => Here we can divide by 4 8 times 6 + 1 ( 16 can be divided 2 times ) + 1 (8 and 24 divided by 4 gives 2 and 6 which is again by 4 an another time)

2 6 10 14 18 22 26 it is all multiple of 2 but not 4 thus we need to make a pair to divide them by 4 (ex: 2*6=12 => Divided by 4) => Here we have 3 pairs possibles so we can divided 3 times

8 + 3 = 11 => D


LHC8717

what if question is to find 4^n in 200! ??
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Re: What is the greatest value of integer n such that 4^n is a factor of 2  [#permalink]

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New post 12 Jan 2017, 03:57
It took me around 1 min 30 to 2 min, but with a number like 200! Your technique will be definitely faster, but I found it easier for me to have a picture of that.
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Re: What is the greatest value of integer n such that 4^n is a factor of 2  [#permalink]

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New post 12 Jan 2017, 08:53
First I solved using below method. From 26!
24 = 4x6
20 = 4x5
16 = 4x4
12 = 4x3
8 = 4x2
4 = 4x1
So n = 7
Why this is wrong? Thanks
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Re: What is the greatest value of integer n such that 4^n is a factor of 2  [#permalink]

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New post 12 Jan 2017, 09:15
kanusha wrote:
First I solved using below method. From 26!
24 = 4x6
20 = 4x5
16 = 4x4
12 = 4x3
8 = 4x2
4 = 4x1
So n = 7
Why this is wrong? Thanks


You forgot all the multiple of 2, remember that a number divided by 4 it's the same than divided by 2 twice

For example you have 2 and 6 in 26! 2*6 = 12 it's divided by 4 (equal 3)

I also highlight in red the remaining multiple of 2 (8 could be divided by 4 once and after by 2 = 1)
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Re: What is the greatest value of integer n such that 4^n is a factor of 2  [#permalink]

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New post 12 Jan 2017, 10:06
rohit8865 wrote:
Bunuel wrote:
What is the greatest value of integer n such that 4^n is a factor of 26! ?

A. 6
B. 9
C. 10
D. 11
E. 12



no. of 2's in 26!
26/2=13
26/4=6
26/8=3
26/16=1

total 13+6+3+1=23
thus no of 4's(4^n)=23/2=11

Ans D


Hi. Great method.
Do we just take the closest integer to be multiplied by? For example: 23/2= 11.5 which is approximately 12 but you took 11. We do not have to round off?
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Re: What is the greatest value of integer n such that 4^n is a factor of 2  [#permalink]

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New post 12 Jan 2017, 18:26
Shiv2016 wrote:
rohit8865 wrote:
Bunuel wrote:
What is the greatest value of integer n such that 4^n is a factor of 26! ?

A. 6
B. 9
C. 10
D. 11
E. 12



no. of 2's in 26!
26/2=13
26/4=6
26/8=3
26/16=1

total 13+6+3+1=23
thus no of 4's(4^n)=23/2=11

Ans D


Hi. Great method.
Do we just take the closest integer to be multiplied by? For example: 23/2= 11.5 which is approximately 12 but you took 11. We do not have to round off?


Shiv2016

yes 11 will be cosidered and rremainder is ignored
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Re: What is the greatest value of integer n such that 4^n is a factor of 2  [#permalink]

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New post 16 Jan 2017, 17:27
2
1
Bunuel wrote:
What is the greatest value of integer n such that 4^n is a factor of 26! ?

A. 6
B. 9
C. 10
D. 11
E. 12



Since 4 = 2^2, we are actually trying to determine the largest value of n such that 2^(2n) is a factor of 26!.

Let’s first determine the number of factors of 2 within 26!. To do that, we can use the following shortcut in which we divide 26 by 2, and then divide the quotient of 26/2 by 2 and continue this process until we can no longer get a nonzero integer as the quotient.

26/2 = 13

13/2 = 6 (we can ignore the remainder)

6/2 = 3

3/2 = 1 (we can ignore the remainder)

Since 1/2 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 2 within 26!.

Thus, there are 13 + 6 + 3 + 1 = 23 factors of 2 within 26!

However, we are not asked for the number of factors of 2; instead we are asked for the number of factors of 4. Since there are 23 factors of 2 within 26!, and since 23/2 = 11 remainder 1, there are 11 factors of 4 with a factor of 2 left over.

In other words, 2^23 = 4^11 x 2 (which equals 2^22 x 2^1 = 2^23).

Answer: D
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Re: What is the greatest value of integer n such that 4^n is a factor of 2  [#permalink]

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Re: What is the greatest value of integer n such that 4^n is a factor of 2 &nbs [#permalink] 17 Sep 2018, 16:11
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