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

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

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New post 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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Re: non -conventional method reqd  [#permalink]

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New post 05 Sep 2008, 01:12
Is this n=(2^4)*3*5, or n=2^(4*3*5) ?
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Re: non -conventional method reqd  [#permalink]

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New post 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?
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Re: non -conventional method reqd  [#permalink]

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New post 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.
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Re: non -conventional method reqd  [#permalink]

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New post 25 Oct 2008, 02:26
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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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Re: non -conventional method reqd  [#permalink]

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New post 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.
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Re: non -conventional method reqd  [#permalink]

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New post 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 :)
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Re: non -conventional method reqd  [#permalink]

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

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New post 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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Re: If n=2^4*3*5, how many factors of n are greater than or  [#permalink]

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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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