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# n*k denotes the product of all integers from n to k,

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Manager
Joined: 04 Jan 2008
Posts: 118
n*k denotes the product of all integers from n to k, [#permalink]

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12 Sep 2008, 06:39
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100% (02:51) correct 0% (00:00) wrong based on 2 sessions

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n*k denotes the product of all integers from n to k, inclusive. If (92*103)/$$6^M$$ is an integer, what is the maximum value of M ?

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SVP
Joined: 07 Nov 2007
Posts: 1799
Location: New York

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12 Sep 2008, 21:46
dancinggeometry wrote:
n*k denotes the product of all integers from n to k, inclusive. If (92*103)/$$6^M$$ is an integer, what is the maximum value of M ?

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6=2*3
We know every alternate number is even.and divisable by 2..
we have to find out how many integers are divisable by 3
103-92 --> 12 numbers present 12/3=4 4 integers are divisable by 3

so max M=4
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Joined: 12 Jul 2008
Posts: 515
Schools: Wharton

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12 Sep 2008, 21:54
dancinggeometry wrote:
n*k denotes the product of all integers from n to k, inclusive. If (92*103)/$$6^M$$ is an integer, what is the maximum value of M ?

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5
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You need to prime factorize each inetger from n to k, looking for 2s and 3s to see how many 6s you can make.

There will be less 3s than 2s, so just count the number of 3s.

93: 1
96: 1
99: 2
102: 1

B
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Joined: 07 Nov 2007
Posts: 1799
Location: New York

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12 Sep 2008, 21:58
zoinnk wrote:
dancinggeometry wrote:
n*k denotes the product of all integers from n to k, inclusive. If (92*103)/$$6^M$$ is an integer, what is the maximum value of M ?

4
5
8
11
12

You need to prime factorize each inetger from n to k, looking for 2s and 3s to see how many 6s you can make.

There will be less 3s than 2s, so just count the number of 3s.

93: 1
96: 1
99: 2
102: 1

B

opps missed that!!!
Thanks zoinnk
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SVP
Joined: 07 Nov 2007
Posts: 1799
Location: New York

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13 Sep 2008, 11:39
hi -

Why 99:2?

zoinnk wrote:
dancinggeometry wrote:
n*k denotes the product of all integers from n to k, inclusive. If (92*103)/$$6^M$$ is an integer, what is the maximum value of M ?

4
5
8
11
12

You need to prime factorize each inetger from n to k, looking for 2s and 3s to see how many 6s you can make.

There will be less 3s than 2s, so just count the number of 3s.

93: 1
96: 1
99: 2
102: 1

B

99 is divisible by 9 (i.e 3*3) so there are two threes
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VP
Joined: 17 Jun 2008
Posts: 1381

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14 Sep 2008, 00:57
hi -

Why 99:2?

zoinnk wrote:
dancinggeometry wrote:
n*k denotes the product of all integers from n to k, inclusive. If (92*103)/$$6^M$$ is an integer, what is the maximum value of M ?

4
5
8
11
12

You need to prime factorize each inetger from n to k, looking for 2s and 3s to see how many 6s you can make.

There will be less 3s than 2s, so just count the number of 3s.

93: 1
96: 1
99: 2
102: 1

B

99 is divisible by 9 (i.e 3*3) so there are two threes

IMO B
same approach look for 3 since that decide 6 formed
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Re: Zumit PS 013   [#permalink] 14 Sep 2008, 00:57
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