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# What is the greatest value of m such that 4m is a factor of

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Manager
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What is the greatest value of m such that 4m is a factor of [#permalink]  25 Jun 2006, 17:11
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What is the greatest value of m such that 4m is a factor of 30! ?

(A) 13
(B) 12
(C) 11
(D) 7
(E) 6
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Re: PS : Exponents [#permalink]  25 Jun 2006, 18:10
briozeal wrote:
What is the greatest value of m such that 4m is a factor of 30! ?

(A) 13
(B) 12
(C) 11
(D) 7
(E) 6

i guess 4m=4^m.

go with A. 13.

2=30/2=15
4=30/4=7
8=30/8=3
16=30/16=1
total 2's = 26 i.e thirteen (13) 4's.
Manager
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Ans is 13 if the expression is 4^m. I don't think 4m is the right expression.
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Thanks,
Zooroopa

Manager
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Re: PS : Exponents [#permalink]  25 Jun 2006, 23:08
MA wrote:
go with A. 13.

2=30/2=15
4=30/4=7
8=30/8=3
16=30/16=1
total 2's = 26 i.e thirteen (13) 4's.

I know I am dumb But I would be very thankful if somebody explain this a little bit more, as I wasn't able to get why we divided here only upto 16 or what exactly is logic behind this division.
VP
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Re: PS : Exponents [#permalink]  26 Jun 2006, 13:27
Quote:
What is the greatest value of m such that 4^m is a factor of 30! ?

(A) 13
(B) 12
(C) 11
(D) 7
(E) 6

4^m = 2^2m
30! = 1x2x3x4x5x6x7x8x9x10x......................x30.
if we count on factors of 30!, we have

1. 15 two's.
2. 7 four's.
3. 3 eight's
4. 1 sixteens.

this is because if we only count 2's, we ignore some other 2's. for ex: 4 has 2 two's. therefore, we do this procedure to find the number of 2's or 4's.

if we add all 2's, we have 26 two's (2's).
2^26 = 4^13.

i am not sure whether i explained it clearly. I will try again if any.

humans wrote:
MA wrote:
go with A. 13.

2=30/2=15
4=30/4=7
8=30/8=3
16=30/16=1
total 2's = 26 i.e thirteen (13) 4's.

I know I am dumb But I would be very thankful if somebody explain this a little bit more, as I wasn't able to get why we divided here only upto 16 or what exactly is logic behind this division.
Director
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The answer is A. To break down any factorial into its prime factors, here is what you do:-

4^m = 2^2m

So we need to find the highest m for which 2^2m divides 30!.

30/2 = 15 (take only the quotient)
15/2 = 7 (take only the quotient)
7/2 = 3 (take only the quotient)
3/2 = 1 (take only the quotient)

When you get a quotient that is not divisible by the divisor or will give a fraction, stop. Add all the quotients, and you have the highest power of the number 2 that divides 30!. As you can see, this is 15 + 7 + 3 + 1 = 26.

Therefore 2m=26 or m=13.

Hope this helps.

Now if the question said 4m divides 30!, you just need to find the highest number that divides 30!/4, which is again 13.

Manager
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Futuristic wrote:
The answer is A. To break down any factorial into its prime factors, here is what you do:-

4^m = 2^2m

So we need to find the highest m for which 2^2m divides 30!.

30/2 = 15 (take only the quotient)
15/2 = 7 (take only the quotient)
7/2 = 3 (take only the quotient)
3/2 = 1 (take only the quotient)

When you get a quotient that is not divisible by the divisor or will give a fraction, stop. Add all the quotients, and you have the highest power of the number 2 that divides 30!. As you can see, this is 15 + 7 + 3 + 1 = 26.

Therefore 2m=26 or m=13.

Hope this helps.

Now if the question said 4m divides 30!, you just need to find the highest number that divides 30!/4, which is again 13.

sorry in advance if this is a dumb question, but I understand your explanation, but does this alway work? Do I always have to bring it to a base of 2 and then do what you did? What happens if it was not 4^m, but 5^m? Thanks
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