Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...