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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.
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.
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.
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 ! _________________
ash
________________________
I'm crossing the bridge.........
Re: PS Combinations [#permalink]
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.
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 !
Re: PS Combinations [#permalink]
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. _________________
Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.
Re: PS Combinations [#permalink]
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.
Re: PS Combinations [#permalink]
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.
Re: PS Combinations [#permalink]
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.
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. _________________
Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.
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.
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