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# What is the greatest value of x such that 8^x is a factor of

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What is the greatest value of x such that 8^x is a factor of [#permalink]

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17 Oct 2008, 07:27
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1. What is the greatest value of $$x$$ such that $$8^x$$ is a factor of 16!?
(A) 2
(B) 3
(C) 5
(D) 6
(E) 8

2. What is the greatest value of q such that 9^q is a factor of 21! ?
(A) 1
(B) 3
(C) 4
(D) 5
(E) 6

Please show me the fastest way to approach this.

Thanks

Last edited by tarek99 on 23 Oct 2008, 09:30, edited 1 time in total.
Senior Manager
Joined: 21 Apr 2008
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Re: PS: greatest value of x [#permalink]

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17 Oct 2008, 13:05
C = 5

This is how approached it

8^x = 2^(3x)
From 16! pick all the number that can be divided by 2

16*8*4*2 = 2^10
and
14*12*10*6 = 2^5

so 2^(3x) = 2^15
x=5

I am sure there are formulae to solve it more easily, but this is what I could do
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Re: PS: greatest value of x [#permalink]

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21 Oct 2008, 06:02
Here's a better approach...

Combination of 4 and 2 will give a product of eight. So if we find the max. no. of pairs of 4 and 2 we can find the max. of x.

Now the no. of 4's that we can extract out of 16! is less than the the no. of 2's. In other words 2's are available in plenty in comparison to 4's. So our answer gets further restricted to the max. no.s of of 4's.

Now the no. of 4's is given by the 16/4 = 4 and 16/ 4^2 =1

therefore 5
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Thanks
rampuria

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Re: PS: greatest value of x [#permalink]

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22 Oct 2008, 02:38
rampuria wrote:
Here's a better approach...

Combination of 4 and 2 will give a product of eight. So if we find the max. no. of pairs of 4 and 2 we can find the max. of x.

Now the no. of 4's that we can extract out of 16! is less than the the no. of 2's. In other words 2's are available in plenty in comparison to 4's. So our answer gets further restricted to the max. no.s of of 4's.

Now the no. of 4's is given by the 16/4 = 4 and 16/ 4^2 =1

therefore 5

Agreed
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"You have to find it. No one else can find it for you." - Bjorn Borg

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Re: PS: greatest value of x [#permalink]

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22 Oct 2008, 09:32
My approach was to figure out how many 8's I can make out of 16!

16 = 8 X 2 14 = 2 X 7 and so on for all even numbers.

8 can also be arrived at by 2 ^ 3 or 4 X 2 ( such numbers are formed from left out of numbers of 14 , 12 etc)

we get 5 eights and hence x=5
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Re: PS: greatest value of x [#permalink]

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22 Oct 2008, 16:48
rampuria wrote:
Here's a better approach...

Combination of 4 and 2 will give a product of eight. So if we find the max. no. of pairs of 4 and 2 we can find the max. of x.

Now the no. of 4's that we can extract out of 16! is less than the the no. of 2's. In other words 2's are available in plenty in comparison to 4's. So our answer gets further restricted to the max. no.s of of 4's.

Now the no. of 4's is given by the 16/4 = 4 and 16/ 4^2 =1

therefore 5

so according to your approach, how would you then solve this problem:

What is the greatest value of $$q$$ such that $$9^q$$ is a factor of 21! ?
(A) 1
(B) 3
(C) 4
(D) 5
(E) 6
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Re: PS: greatest value of x [#permalink]

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22 Oct 2008, 19:28
LiveStronger wrote:
C - 4

3^8 = 3^2q
q = 4

for whatever reason i m getting 3^9 = 3^2q
Pls help me see why ?

3 6 9 12 15 18 21 = 3^9
1+1+2+1+1+2+1=9
Re: PS: greatest value of x   [#permalink] 22 Oct 2008, 19:28
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