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Sets, Progressions, Combinations - 1

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Sets, Progressions, Combinations - 1

by coolhabhi » Sun Feb 26, 2012 10:40 pm
Given a set
A = {1, 3, 9, 27, 81, -----, (3)^100}

Y is a subset of A such that the GEOMETRIC mean of no two numbers is (3)^50. N is the maximum possible number of elements in set Y.

1) What is the value of N?
A) 50
B) 51
C) 49
D) 52

2) In how many ways can the set Y be formed such that it has exactly N elements?
A) (2)^50
B) (2)^51
C) (3)^50
D) (3)^51

Source: A test booklet which I recently found.
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by sanju09 » Mon Feb 27, 2012 4:54 am
coolhabhi wrote:Given a set
A = {1, 3, 9, 27, 81, -----, (3)^100}

Y is a subset of A such that the GEOMETRIC mean of no two numbers is (3)^50. N is the maximum possible number of elements in set Y.

1) What is the value of N?
A) 50
B) 51
C) 49
D) 52

2) In how many ways can the set Y be formed such that it has exactly N elements?
A) (2)^50
B) (2)^51
C) (3)^50
D) (3)^51

Source: A test booklet which I recently found.

The geometric mean of two positive numbers a and b is √ (ab), and if a and b are taken from the set {1, 3, 9, 27, 81, -----, (3)^100}, then if set Y contains 1, it won't contain (3)^100; if set Y contains 3, it won't contain (3)^99; if set Y contains (3)^2, it won't contain (3)^98... if set Y contains (3)^49, it won't contain (3)^51, and if set Y contains (3)^50, then there is no issue as we can't have another (3)^50 in the same set. Hence, if 1, 3, (3)^2, ...(3)^50 are in Y, then (3)^100, (3)^99, (3)^98, ...(3)^51 are not in Y. That way, the maximum possible cardinal number of any Y is [spoiler]51. Or the maximum N = 51.

Take B[/spoiler]
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by Brent@GMATPrepNow » Mon Feb 27, 2012 8:46 am
I should point out that the GMAT does not require any knowledge of Geometric Mean.
One way to rephrase the opening part of question so that it's appropriate for the GMAT would be as follows:

A = {1, 3, 9, 27, 81, -----, (3)^100}

Y is a subset of A. No two numbers in Y have a product of (3)^100. N is the maximum possible number of elements in set Y.

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by Brent@GMATPrepNow » Mon Feb 27, 2012 9:02 am
Let's answer these questions 1 at a time (with the GMAT-appropriate wording)
coolhabhi wrote:Given a set
A = {1, 3, 9, 27, 81, -----, (3)^100}

Y is a subset of A. No two numbers in Y have a product of (3)^100. N is the maximum possible number of elements in set Y.

1) What is the value of N?
A) 50
B) 51
C) 49
D) 52
First, let's rewrite set A as: {3^0, 3^1, 3^2, 3^3, . . . 3^98, 3^99, 3^100}

Notice that there are 101 numbers in set A. Also notice that 50 of the values are less than 3^50 and 50 of the values are greater than 3^50.

We have {3^0, 3^1, 3^2, 3^3,...3^49, 3^50, 3^51..., 3^98, 3^99, 3^100}

Now subset Y cannot have any pair of numbers with a product of 3^100.
So, for example, set Y cannot include the pair of numbers 3^0 and 3^100, since their product is 3^100.
Similarly, set Y cannot include the pair of numbers 3^2 and 3^98, since their product is 3^100.

Big point: For every number less than 3^50, there's a corresponding number greater than 3^50 such that the product of the two values is 3^100.

So, we essentially have 2 pairs of values (one less than 3^50 and one greater than 3^50) such that the product of the two values is 3^100.

This means that subset A can have (at most) only one value from each of these pairs. So, if we take one value from each pair, then we have 50 numbers in subset A.

Also notice that we have have 3^50 in subset A since it does not have a corresponding (different) value such that the product is 3^100.

So, the maximum number of values (N) in subset A is [spoiler]51 (B)[/spoiler].

Cheers,
Brent
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by Brent@GMATPrepNow » Mon Feb 27, 2012 9:17 am
Now let's answer the second questions (with the GMAT-appropriate wording)
coolhabhi wrote:Given a set
A = {1, 3, 9, 27, 81, -----, (3)^100}

Y is a subset of A. No two numbers in Y have a product of (3)^100. N is the maximum possible number of elements in set Y.

2) In how many ways can the set Y be formed such that it has exactly N elements?
A) (2)^50
B) (2)^51
C) (3)^50
D) (3)^51
Well, to begin, we know (from part 1) that N=51. So, in how many ways can set Y be formed such that it has exactly 51 elements?

Well, remember that set A can have 1 number from the following pairs of values.
(3^0 or 3^100)
(3^1 or 3^99)
(3^2 or 3^98)
.
.
.
(3^49 or 3^51)

There are 50 pairs of values. And we must select one value from each set.
Well, we can select one value from the 1st set in 2 ways.
We can select one value from the 2nd set in 2 ways.
We can select one value from the 3rd set in 2 ways.
.
.
.
We can select one value from the 50th set in 2 ways.

So, the total number of ways to select one number from each set = (2)(2)(2)...(2) = 2^50

(Aside: we must also include 3^50 in set A, but there's only 1 way to select 3^50 to be in the set, so it doesn't change our calculations)

So, the final answer is [spoiler]2^50 - A[/spoiler]

Cheers,
Brent
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by voodoo_child » Fri Mar 09, 2012 7:04 am
Brent@GMATPrepNow wrote:Let's answer these questions 1 at a time (with the GMAT-appropriate wording)
coolhabhi wrote:Given a set
A = {1, 3, 9, 27, 81, -----, (3)^100}

Y is a subset of A. No two numbers in Y have a product of (3)^100. N is the maximum possible number of elements in set Y.

1) What is the value of N?
A) 50
B) 51
C) 49
D) 52
First, let's rewrite set A as: {3^0, 3^1, 3^2, 3^3, . . . 3^98, 3^99, 3^100}

Notice that there are 101 numbers in set A. Also notice that 50 of the values are less than 3^50 and 50 of the values are greater than 3^50.

We have {3^0, 3^1, 3^2, 3^3,...3^49, 3^50, 3^51..., 3^98, 3^99, 3^100}

Now subset Y cannot have any pair of numbers with a product of 3^100.
So, for example, set Y cannot include the pair of numbers 3^0 and 3^100, since their product is 3^100.
Similarly, set Y cannot include the pair of numbers 3^2 and 3^98, since their product is 3^100.

Big point: For every number less than 3^50, there's a corresponding number greater than 3^50 such that the product of the two values is 3^100.

So, we essentially have 2 pairs of values (one less than 3^50 and one greater than 3^50) such that the product of the two values is 3^100.

This means that subset A can have (at most) only one value from each of these pairs. So, if we take one value from each pair, then we have 50 numbers in subset A.

Also notice that we have have 3^50 in subset A since it does not have a corresponding (different) value such that the product is 3^100.

So, the maximum number of values (N) in subset A is [spoiler]51 (B)[/spoiler].

Cheers,
Brent
Brent - I have a question on your post:

I am completely off on this one. Here's what I did:

Two numbers can be chosen in 101C2 ways = 101*50

Now, there are 50 cases where 3^X = X^100

Therefore total possibilities = 101*50 - 50 = 100*50.

Any help?
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by Brent@GMATPrepNow » Fri Mar 09, 2012 7:43 am
voodoo_child wrote:
Brent - I have a question on your post:

I am completely off on this one. Here's what I did:

Two numbers can be chosen in 101C2 ways = 101*50

Now, there are 50 cases where 3^X = X^100

Therefore total possibilities = 101*50 - 50 = 100*50.

Any help?
I'm not sure what you mean by: there are 50 cases where 3^X = X^100

Can you please explain?

Cheers,
Brent
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by voodoo_child » Fri Mar 09, 2012 12:29 pm
Brent@GMATPrepNow wrote:
voodoo_child wrote:
Brent - I have a question on your post:

I am completely off on this one. Here's what I did:

Two numbers can be chosen in 101C2 ways = 101*50

Now, there are 50 cases where 3^X = X^100

Therefore total possibilities = 101*50 - 50 = 100*50.

Any help?
I'm not sure what you mean by: there are 50 cases where 3^X = X^100

Can you please explain?

Cheers,
Brent
I am sorry, Brent. I meant : 3^X1 * 3^X2 = 3^100; where X1 and X2 are the two exponents that were picked.

It is similar to what you have mentioned in your solution. i.e. 3^0* 3^100; 3^1 * 3^99 i.e. a pair that when multiplied equal 3^100.



Thanks
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by Brent@GMATPrepNow » Fri Mar 09, 2012 1:41 pm
voodoo_child wrote:
Brent@GMATPrepNow wrote:
voodoo_child wrote:
Brent - I have a question on your post:

I am completely off on this one. Here's what I did:

Two numbers can be chosen in 101C2 ways = 101*50

Now, there are 50 cases where 3^X = X^100

Therefore total possibilities = 101*50 - 50 = 100*50.

Any help?
I'm not sure what you mean by: there are 50 cases where 3^X = X^100

Can you please explain?

Cheers,
Brent
I am sorry, Brent. I meant : 3^X1 * 3^X2 = 3^100; where X1 and X2 are the two exponents that were picked.

It is similar to what you have mentioned in your solution. i.e. 3^0* 3^100; 3^1 * 3^99 i.e. a pair that when multiplied equal 3^100.

Thanks
I see.
Well, there are several problems with your solution.
To begin, you set out by determining the total number of ways to select two values (101C2 or 5050 ways).
So, there are 5050 different pairs of values. Why are you counting the number of pairs? The question isn't asking you to count pairs. It asks you to find the greatest number of elements in a subset.

Also, since there are 101 elements in the original set, the subset cannot contain more than 101 elements (as your solution suggests).

Cheers,
Brent
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