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# If n=2^4*3*5, how many factors of n are greater than or

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Joined: 11 May 2008
Posts: 543
If n=2^4*3*5, how many factors of n are greater than or  [#permalink]

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05 Sep 2008, 00:45
1
. If n=2^4*3*5, how many factors of n are greater than or equal to 8 and less than or equal to 30?
(A) 8
(B) 9
(C) 10
(D) 12
(E) 15

what is the fastest way to solve this?

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Senior Manager
Joined: 16 Jul 2008
Posts: 278
Re: non -conventional method reqd  [#permalink]

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05 Sep 2008, 01:12
Is this n=(2^4)*3*5, or n=2^(4*3*5) ?
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Current Student
Joined: 11 May 2008
Posts: 543
Re: non -conventional method reqd  [#permalink]

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05 Sep 2008, 03:40
alpha_plus_gamma wrote:
arjtryarjtry wrote:
. If n=2^4*3*5, how many factors of n are greater than or equal to 8 and less than or equal to 30?
(A) 8
(B) 9
(C) 10
(D) 12
(E) 15

what is the fastest way to solve this?

Is it A . 8

i know... but could you pls suggest if u used any other method?
VP
Joined: 21 Jul 2006
Posts: 1361
Re: non -conventional method reqd  [#permalink]

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05 Sep 2008, 04:01
1
arjtryarjtry wrote:
alpha_plus_gamma wrote:
arjtryarjtry wrote:
. If n=2^4*3*5, how many factors of n are greater than or equal to 8 and less than or equal to 30?
(A) 8
(B) 9
(C) 10
(D) 12
(E) 15

what is the fastest way to solve this?

Is it A . 8

i know... but could you pls suggest if u used any other method?

I got A as well, but I used the manual way which is:

1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240

Here's what I did, start with 1 on the left side of the scale, with 1, you will need to multiply it by 240, which is on the other end of the scale in order to get the product of 240. Next, find the next number higher than 1 that can divide 240. That second number is 2, so with 2, you will need to multiply it by 120, which is on the other end of the scale, in order to get the product of 240. Next, you will need a 3, so you will need to multiply it by 80, which is on the other end of the scale, in order to get the product of 240. So as you keep doing this, you will have listed out all the numbers that are factors of 240. When you're done with your list, just count the numbers that are at least 8 and at most 30, their number is 8.

Any other faster way will be welcomed! But with my approach, it took me less than a minute to list this down.
Director
Joined: 30 Jun 2008
Posts: 968
Re: non -conventional method reqd  [#permalink]

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25 Oct 2008, 02:26
1
tarek99 wrote:

I got A as well, but I used the manual way which is:

1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240

Here's what I did, start with 1 on the left side of the scale, with 1, you will need to multiply it by 240, which is on the other end of the scale in order to get the product of 240. Next, find the next number higher than 1 that can divide 240. That second number is 2, so with 2, you will need to multiply it by 120, which is on the other end of the scale, in order to get the product of 240. Next, you will need a 3, so you will need to multiply it by 80, which is on the other end of the scale, in order to get the product of 240. So as you keep doing this, you will have listed out all the numbers that are factors of 240. When you're done with your list, just count the numbers that are at least 8 and at most 30, their number is 8.

Any other faster way will be welcomed! But with my approach, it took me less than a minute to list this down.

That was neat Tarek !!

Does anyone know a faster approach to this problem ??

Thanks
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VP
Joined: 17 Jun 2008
Posts: 1357
Re: non -conventional method reqd  [#permalink]

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25 Oct 2008, 04:33
1
Not sure if my approach is faster.

8 = 2^3
30 = 2*3*5

Hence, any factor that is greater than or equal to 8 and less than or equal to 2*3*5 will be
2^3
2^4
2^3*3
2^4*3
2^3*5
2^4*5
2^3*3*5
2^4*3*5

Hence, a total of 8.
Senior Manager
Joined: 21 Apr 2008
Posts: 256
Location: Motortown
Re: non -conventional method reqd  [#permalink]

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25 Oct 2008, 13:20
1
scthakur wrote:
Not sure if my approach is faster.

8 = 2^3
30 = 2*3*5

Hence, any factor that is greater than or equal to 8 and less than or equal to 2*3*5 will be
2^3
2^4
2^3*3
2^4*3
2^3*5
2^4*5
2^3*3*5
2^4*3*5

Hence, a total of 8.

This is how I did it too
Director
Joined: 30 Jun 2008
Posts: 968
Re: non -conventional method reqd  [#permalink]

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26 Oct 2008, 00:59
I guess there is no faster method for this ....
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Posts: 1361
Re: non -conventional method reqd  [#permalink]

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26 Oct 2008, 02:29
amitdgr wrote:
tarek99 wrote:

I got A as well, but I used the manual way which is:

1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240

Here's what I did, start with 1 on the left side of the scale, with 1, you will need to multiply it by 240, which is on the other end of the scale in order to get the product of 240. Next, find the next number higher than 1 that can divide 240. That second number is 2, so with 2, you will need to multiply it by 120, which is on the other end of the scale, in order to get the product of 240. Next, you will need a 3, so you will need to multiply it by 80, which is on the other end of the scale, in order to get the product of 240. So as you keep doing this, you will have listed out all the numbers that are factors of 240. When you're done with your list, just count the numbers that are at least 8 and at most 30, their number is 8.

Any other faster way will be welcomed! But with my approach, it took me less than a minute to list this down.

That was neat Tarek !!

Does anyone know a faster approach to this problem ??

Thanks

thanks
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Joined: 09 Sep 2013
Posts: 11013
Re: If n=2^4*3*5, how many factors of n are greater than or  [#permalink]

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09 Jul 2017, 22:29
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This discussion does not meet community quality standards. It has been retired.

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Re: If n=2^4*3*5, how many factors of n are greater than or   [#permalink] 09 Jul 2017, 22:29
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