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Director
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How many different numbers can be created when multiplying 1 [#permalink]
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27 May 2005, 01:40
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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}?



Manager
Joined: 28 Aug 2004
Posts: 205

any answer choices?
= 6+6^2+6^3+6^4+6^5+6^6
= 6*(6^(n1)) where n is the number of digits
I hope I am not hallucinating!



Manager
Joined: 08 Mar 2005
Posts: 99

I got a 63.
Will explain if its correct...



Intern
Joined: 14 Oct 2004
Posts: 18
Location: Orange County, CA

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.



Director
Joined: 18 Apr 2005
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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, 15:30, edited 1 time in total.



Director
Joined: 18 Apr 2005
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Location: Canuckland

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



Director
Joined: 18 Apr 2005
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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.



Senior Manager
Joined: 21 Mar 2004
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Location: Cary,NC

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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Joined: 17 Apr 2005
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Re: PS Combinations [#permalink]
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29 May 2005, 11:41
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.



Director
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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]
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29 May 2005, 22:40
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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Director
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Re: PS Combinations [#permalink]
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30 May 2005, 02:50
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]
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30 May 2005, 06:39
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]
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30 May 2005, 12:02
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+2^32^2
So total would be 1+2^4+2*2^32^2=29. Wow.
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Joined: 03 Jan 2005
Posts: 2233

Actually, I just got sparky's solution. That is better than what I did. Total numbers of numbers using 25 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+2^52^2=29.
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Director
Joined: 18 Apr 2005
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I just realized that 3*4 = 2*6 =12
so need to subtruct an additional 2
boy, this problem is a pain



Senior Manager
Joined: 21 Mar 2004
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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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Joined: 03 Jan 2005
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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?
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Keep on asking, and it will be given you;
keep on seeking, and you will find;
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Director
Joined: 18 Apr 2005
Posts: 545
Location: Canuckland

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



Director
Joined: 18 Apr 2005
Posts: 545
Location: Canuckland

Unless anyone has any objections and I didn't make a mistake again, let's consider the case closed
And I hope you guys enjoyed solvingit as much as I did
Hopefully, later I will try to find something even more wicked than this problem







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