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How many different numbers can be created when multiplying 1

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How many different numbers can be created when multiplying 1 [#permalink] New post 27 May 2005, 00:40
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How many different numbers can be created when multiplying 1 by any group of numbers from {1,2,3,4,5,6}?
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 [#permalink] New post 27 May 2005, 07:35
any answer choices?

= 6+6^2+6^3+6^4+6^5+6^6

= 6*(6^(n-1)) where n is the number of digits

I hope I am not hallucinating! :-D
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 [#permalink] New post 27 May 2005, 08:56
I got a 63.

Will explain if its correct... :-)
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 [#permalink] New post 27 May 2005, 10:33
This question is a bit confusing. Do you have to use all 6 numbers, or can you use 1 number, or 2 numbers out of that list and so on?

If you have to use 6 numbers at a time my answer is 720. If you can use any number combination from 1 to 6 numbers my answer is 1951. I could be totally off base because I don't really understand the question.

:?
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 [#permalink] New post 27 May 2005, 13:39
KristinHR wrote:
This question is a bit confusing. Do you have to use all 6 numbers, or can you use 1 number, or 2 numbers out of that list and so on?

If you have to use 6 numbers at a time my answer is 720. If you can use any number combination from 1 to 6 numbers my answer is 1951. I could be totally off base because I don't really understand the question.

:?


The question said any group of numbers so it can be1*1 or 1*1*2*3*4*5*6, for example

and there are no repeats like using 6 twice (6*6) because you need to use numbers strictly from the set.

This question was quite clear to me :)

Last edited by sparky on 27 May 2005, 14:30, edited 1 time in total.
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 [#permalink] New post 27 May 2005, 14:10
My answer to this question is

2^5 - 2^2 = 32 - 4 = 28 (number of different numbers that can be created, ranging from 1 to 6!)

you need to subtruct 2^2 because 2*3 = 6, to eliminate doublecounting
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 [#permalink] New post 29 May 2005, 00:08
There are two options for each number in the set {1,2,3,4,5,6} to be included in the group or to be excluded from the group. There are 6 numbers in the group. 1 doesn't count since multiplying any number by 1 doesn't change the number. So, the number of products that result in different numbers is 2^5, 2*3, 2*3*4, 3*4, 3*5*6, etc . Subtruct 2^2 from it since 2*3=6.
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 [#permalink] New post 29 May 2005, 08:50
sparky wrote:
There are two options for each number in the set {1,2,3,4,5,6} to be included in the group or to be excluded from the group. There are 6 numbers in the group. 1 doesn't count since multiplying any number by 1 doesn't change the number. So, the number of products that result in different numbers is 2^5, 2*3, 2*3*4, 3*4, 3*5*6, etc . Subtruct 2^2 from it since 2*3=6.



Sparky ,
I understood the 2^5 part but not the 2^2 part.

2*3 = 6

3*4 = 12
2*6 = 12

3*4*5 = 60
2*6*5 = 60

2*3*4 = 24
6*4 = 24

2*3*5 = 30
6*5 = 30

How do you explain 2^2 now ?

I am not able to figure out any formula to eliminate double counting !
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Re: PS Combinations [#permalink] New post 29 May 2005, 10:41
sparky wrote:
How many different numbers can be created when multiplying 1 by any group of numbers from {1,2,3,4,5,6}?


32.

Remove 6 from the set as it has a factor 2 and 3 repeated.
Remove 1 from the set as it doesn't create any new value(eg . (1,2,3 ) would give the same as (2,3) )

Hece we are left with 4 elements hence 15 different sets are possible
Let us now add 1 as we had removed this earlier.
So total of 16 different values to far.
Now let us use 6 to create a total of 32 different values.

HTMG.


HMTG.
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 [#permalink] New post 29 May 2005, 12:46
ashkg wrote:
sparky wrote:
There are two options for each number in the set {1,2,3,4,5,6} to be included in the group or to be excluded from the group. There are 6 numbers in the group. 1 doesn't count since multiplying any number by 1 doesn't change the number. So, the number of products that result in different numbers is 2^5, 2*3, 2*3*4, 3*4, 3*5*6, etc . Subtruct 2^2 from it since 2*3=6.



Sparky ,
I understood the 2^5 part but not the 2^2 part.

2*3 = 6


3*4 = 12
2*6 = 12

3*4*5 = 60
2*6*5 = 60

2*3*4 = 24
6*4 = 24

2*3*5 = 30
6*5 = 30

How do you explain 2^2 now ?

I am not able to figure out any formula to eliminate double counting !



2*3*4*5 = 4*5*6
2*3*4 = 4*6
2*3*5 = 5*6
2*3 = 6

and everything is multiplied by 1 of course
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Re: PS Combinations [#permalink] New post 29 May 2005, 21:40
sparky wrote:
How many different numbers can be created when multiplying 1 by any group of numbers from {1,2,3,4,5,6}?


Ok, so the first number (1) doesn't do anything in getting a new number except for when it is 1*1. And the last number (6) doesn't do anything except for when it is 2*6, 3*6 or 2*3*6.

The middle for numbers can be included or not included to result in a new number.

So total number of different numbers would be 1 + 2^4+3=20.
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Re: PS Combinations [#permalink] New post 30 May 2005, 01:50
HongHu wrote:
sparky wrote:
How many different numbers can be created when multiplying 1 by any group of numbers from {1,2,3,4,5,6}?


Ok, so the first number (1) doesn't do anything in getting a new number except for when it is 1*1. And the last number (6) doesn't do anything except for when it is 2*6, 3*6 or 2*3*6.

The middle for numbers can be included or not included to result in a new number.

So total number of different numbers would be 1 + 2^4+3=20.


What about

2x3x4x5x6
2x3x5x6
2x3x4x6
3x4x5x6
3x5x6
3x4x6
2x4x5x6
2x5x6
2x4x6

?
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Re: PS Combinations [#permalink] New post 30 May 2005, 05:39
HowManyToGo wrote:
sparky wrote:
How many different numbers can be created when multiplying 1 by any group of numbers from {1,2,3,4,5,6}?


32.

Remove 6 from the set as it has a factor 2 and 3 repeated.
Remove 1 from the set as it doesn't create any new value(eg . (1,2,3 ) would give the same as (2,3) )

Hece we are left with 4 elements hence 15 different sets are possible
Let us now add 1 as we had removed this earlier.
So total of 16 different values to far.
Now let us use 6 to create a total of 32 different values.

HTMG.


HMTG.


I seem to have a mistake while taking 6 back into consideration
I assument that new numbers would be formed with the introduction of 6 , but consider 1*6 = 6 which is already a number ,2*3

So the part upto 16 unique numbers excluding 6 is correct.
Now for 6,it would only make a difference to numbers which already have 6 as a factor.
Hence amond the 16 unique numbers, the no. of numbers which have 6 as a factor is 4( 6,24,30 and 120), so we would get an addition of 4 numbers with the introduction of 6.

That makes it a total of 20.

HMTG.
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Re: PS Combinations [#permalink] New post 30 May 2005, 11:02
sparky wrote:
HongHu wrote:
sparky wrote:
How many different numbers can be created when multiplying 1 by any group of numbers from {1,2,3,4,5,6}?


Ok, so the first number (1) doesn't do anything in getting a new number except for when it is 1*1. And the last number (6) doesn't do anything except for when it is 2*6, 3*6 or 2*3*6.

The middle for numbers can be included or not included to result in a new number.

So total number of different numbers would be 1 + 2^4+3=20.


What about

2x3x4x5x6
2x3x5x6
2x3x4x6
3x4x5x6
3x5x6
3x4x6
2x4x5x6
2x5x6
2x4x6

?


Yes, that's a good point. I was wrong.

So it would be any number that had included 2 or 3 then: 2^2+2^3-2^2

So total would be 1+2^4+2*2^3-2^2=29. Wow.
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 [#permalink] New post 30 May 2005, 11:06
Actually, I just got sparky's solution. That is better than what I did. Total numbers of numbers using 2-5 would be 2^5, and duplicate numbers are choosing 2 and 3 and not 6 (or choosing 6 and not 2 and 3), so only need to consider 4 and 5, which means number of duplicated numbers are 2^2. Add the 1*1 we'll get

1+2^5-2^2=29.
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 [#permalink] New post 30 May 2005, 13:29
I just realized that 3*4 = 2*6 =12

so need to subtruct an additional 2

boy, this problem is a pain :)
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 [#permalink] New post 30 May 2005, 15:30
sparky wrote:
I just realized that 3*4 = 2*6 =12

so need to subtruct an additional 2

boy, this problem is a pain :)


I had pointed this way before.......!!!!!!!!!

what is the OA and someone please explain.
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 [#permalink] New post 30 May 2005, 16:57
:lol: What are we going to do if this question is in the real test?

I suppose try to list them all up may be the best way. Sigh. :p

So the final answer is 27 then? :wink:
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 [#permalink] New post 30 May 2005, 21:41
ash, sorry man, I guess I didn't read the answers carefully. I don't have an OA for this because I made it up. I thought this level would be comparable to high end GMAT math problems.

Looks like 26 = 2^5 - 2^4 - 2^1. Below I used a brute force computational approach to eliminate all doubts. Repeats are marked with V.

I don't think we need to add 1 to the sum because the case 'no numbers chosen' is included in 2^5 and corresponds to choosing just 1.

Product Combination#
1 1
2 2
3 3
4 4
5 5
6 6
6 7 V
8 8
10 9
12 10 V
12 11
15 12
18 13
20 14
24 15 V
30 16 V
24 17
40 18
60 19 V
30 20
36 21
48 22
60 23
72 24
90 25
120 26 V
120 27
360 28
240 29
180 30
144 31
720 32
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 [#permalink] New post 30 May 2005, 22:08
Unless anyone has any objections and I didn't make a mistake again, let's consider the case closed :-D
And I hope you guys enjoyed solvingit as much as I did :roll:
Hopefully, later I will try to find something even more wicked than this problem :twisted:
  [#permalink] 30 May 2005, 22:08
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