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Fresh Meat!!! [#permalink] New post 17 Apr 2013, 05:11
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The next set of PS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers.

1. The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square S and the area of rhombus R?

I. 63
II. 126
III. 252


A. I only
B. II only
C. III only
D. I and III only
E. I, II and III

Solution: fresh-meat-151046-80.html#p1215318


2. Set S contains 7 different letters. How many subsets of set S, including an empty set, contain at most 3 letters?

A. 29
B. 56
C. 57
D. 63
E. 64

Solution: fresh-meat-151046-100.html#p1215323

3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

A. 16
B. 27
C. 31
D. 32
E. 64

Solution: fresh-meat-151046-100.html#p1215329


4. The functions f and g are defined for all the positive integers n by the following rule: f(n) is the number of positive perfect squares less than n and g(n) is the number of primes numbers less than n. If f(x) + g(x) = 16, then x is in the range:

A. 30 < x < 36
B. 30 < x < 37
C. 31 < x < 37
D. 31 < x < 38
E. 32 < x < 38

Solution: fresh-meat-151046-100.html#p1215335


5. Which of the following is a factor of 18!+1?

A. 15
B. 17
C. 19
D. 33
E. 39

Solution: fresh-meat-151046-100.html#p1215338


6. If the least common multiple of a positive integer x, 4^3 and 6^5 is 6^6. Then x can take how many values?

A. 1
B. 6
C. 7
D. 30
E. 36

Solution: fresh-meat-151046-100.html#p1215345


7. The greatest common divisor of two positive integers is 25. If the sum of the integers is 350, then how many such pairs are possible?

A. 1
B. 2
C. 3
D. 4
E. 5

Solution: fresh-meat-151046-100.html#p1215349


8. The product of a positive integer x and 377,910 is divisible by 3,300, then the least value of x is:

A. 10
B. 11
C. 55
D. 110
E. 330

Solution: fresh-meat-151046-100.html#p1215359


9. What is the 101st digit after the decimal point in the decimal representation of 1/3 + 1/9 + 1/27 + 1/37?

A. 0
B. 1
C. 5
D. 7
E. 8

Solution: fresh-meat-151046-100.html#p1215367


10. If x is not equal to 0 and x^y=1, then which of the following must be true?

I. x=1
II. x=1 and y=0
III. x=1 or y=0


A. I only
B. II only
C. III only
D. I and III only
E. None

Solution: fresh-meat-151046-100.html#p1215370


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Re: Fresh Meat!!! [#permalink] New post 27 Feb 2014, 06:22
Bunuel wrote:
10. If x is not equal to 0 and x^y=1, then which of the following must be true?

I. x=1
II. x=1 and y=0
III. x=1 or y=0


A. I only
B. II only
C. III only
D. I and III only
E. None

Notice that if x=-1 and y is any even number, then (-1)^{even}=1, thus none of the options must be true.

Answer: E.


Bunuel.. (II) x=1 andy=0 will give ans 1.. cant we say that this is the solution dat must give Ans "1"? 1^0=1?
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Re: Fresh Meat!!! [#permalink] New post 27 Feb 2014, 06:25
Bunuel wrote:
3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

A. 16
B. 27
C. 31
D. 32
E. 64

Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.

Answer: D.



not including 0.. so we will have 5 integers.. 5c1 5c2 5c3 5c4 5c5.. i m geting ans 31...

I m not geting how did u get this 2^5? m not geting this logic
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Re: Fresh Meat!!! [#permalink] New post 27 Feb 2014, 06:35
Expert's post
sanjoo wrote:
Bunuel wrote:
10. If x is not equal to 0 and x^y=1, then which of the following must be true?

I. x=1
II. x=1 and y=0
III. x=1 or y=0


A. I only
B. II only
C. III only
D. I and III only
E. None

Notice that if x=-1 and y is any even number, then (-1)^{even}=1, thus none of the options must be true.

Answer: E.


Bunuel.. (II) x=1 andy=0 will give ans 1.. cant we say that this is the solution dat must give Ans "1"? 1^0=1?


The question asks which of the following MUST be true, not COULD be true. So, if if x=-1 and y is any even number, then (-1)^{even}=1, thus none of the options MUST be true.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Fresh Meat!!! [#permalink] New post 27 Feb 2014, 06:35
Expert's post
sanjoo wrote:
Bunuel wrote:
3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

A. 16
B. 27
C. 31
D. 32
E. 64

Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.

Answer: D.



not including 0.. so we will have 5 integers.. 5c1 5c2 5c3 5c4 5c5.. i m geting ans 31...

I m not geting how did u get this 2^5? m not geting this logic


Please read the whole thread: fresh-meat-151046-120.html#p1243696
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Fresh Meat!!! [#permalink] New post 17 Mar 2014, 10:48
Bunuel wrote:
9. What is the 101st digit after the decimal point in the decimal representation of 1/3 + 1/9 + 1/27 + 1/37?

A. 0
B. 1
C. 5
D. 7
E. 8

\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}=\frac{333}{999} + \frac{111}{999} + \frac{37}{999} + \frac{27}{999}=\frac{508}{999}=0.508508....

102nd digit will be 8, thus 101st digit will be 0.

Answer: A.


How will one come to know that 508/999 is a repeating decimal. One would stop at 508/999..
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Re: Fresh Meat!!! [#permalink] New post 03 Apr 2014, 07:25
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Bunuel wrote:
4. The functions f and g are defined for all the positive integers n by the following rule: f(n) is the number of perfect squares less than n and g(n) is the number of primes numbers less than n. If f(x) + g(x) = 16, then x is in the range:

A. 30 < x < 36
B. 30 < x < 37
C. 31 < x < 37
D. 31 < x < 38
E. 32 < x < 38

Perfect squares: 1, 4, 9, 16, 25, 36, ..,
Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, ...

If x = 31, then f(31) = 5 and g(31) = 10: f(x) + g(x) = 5 + 10 = 15.
If x = 32, then f(32) = 5 and g(32) = 11: f(x) + g(x) = 5 + 11 = 16.
...
If x = 36, then f(36) = 5 and g(36) = 11: f(x) + g(x) = 5 + 11 = 16.
If x = 37, then f(37) = 6 and g(37) = 11: f(x) + g(x) = 6 + 11 = 17.

Thus x could be 32, 33, 34, 35 or 36: 31<x<37.

Answer: C.


Why is zero not being considered in the count of perfect squares? I realize that it states that n is positive, but if f(n) is defined as perfect squares less than n and zero is less than, why do we not count it?
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Re: Fresh Meat!!! [#permalink] New post 03 Apr 2014, 07:44
Expert's post
MrWallSt wrote:
Bunuel wrote:
4. The functions f and g are defined for all the positive integers n by the following rule: f(n) is the number of perfect squares less than n and g(n) is the number of primes numbers less than n. If f(x) + g(x) = 16, then x is in the range:

A. 30 < x < 36
B. 30 < x < 37
C. 31 < x < 37
D. 31 < x < 38
E. 32 < x < 38

Perfect squares: 1, 4, 9, 16, 25, 36, ..,
Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, ...

If x = 31, then f(31) = 5 and g(31) = 10: f(x) + g(x) = 5 + 10 = 15.
If x = 32, then f(32) = 5 and g(32) = 11: f(x) + g(x) = 5 + 11 = 16.
...
If x = 36, then f(36) = 5 and g(36) = 11: f(x) + g(x) = 5 + 11 = 16.
If x = 37, then f(37) = 6 and g(37) = 11: f(x) + g(x) = 6 + 11 = 17.

Thus x could be 32, 33, 34, 35 or 36: 31<x<37.

Answer: C.


Why is zero not being considered in the count of perfect squares? I realize that it states that n is positive, but if f(n) is defined as perfect squares less than n and zero is less than, why do we not count it?


You are right. The question should read: f(n) is the number of positive perfect squares less than n. Edited.
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Re: Fresh Meat!!! [#permalink] New post 09 Apr 2014, 00:53
Bunuel wrote:
3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

A. 16
B. 27
C. 31
D. 32
E. 64

Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.

Answer: D.


Hi Bunuel,

Is {NULL} a subset of {1,2,3,4,5}?
Because 2^5 also contains {NULL} as one possibility.

Thanks..
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Re: Fresh Meat!!! [#permalink] New post 09 Apr 2014, 02:05
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riskietech wrote:
Bunuel wrote:
3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

A. 16
B. 27
C. 31
D. 32
E. 64

Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.

Answer: D.


Hi Bunuel,

Is {NULL} a subset of {1,2,3,4,5}?
Because 2^5 also contains {NULL} as one possibility.

Thanks..


Yes, an empty set is a subset of all sets.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Fresh Meat!!! [#permalink] New post 17 May 2014, 06:44
Bunuel wrote:
3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

A. 16
B. 27
C. 31
D. 32
E. 64

Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.

Answer: D.


Dear Bunnel

I didnt understand this.

y didnt we take the set {1,2,3,4,5} into consideration and solve like the above qs? where did we get 2^5 from?
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Re: Fresh Meat!!! [#permalink] New post 17 May 2014, 06:51
Expert's post
NGGMAT wrote:
Bunuel wrote:
3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

A. 16
B. 27
C. 31
D. 32
E. 64

Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.

Answer: D.


Dear Bunnel

I didnt understand this.

y didnt we take the set {1,2,3,4,5} into consideration and solve like the above qs? where did we get 2^5 from?


What do you mean by the red part?

As for 2^5: the number of subsets of n-element set is 2^n, thus the number of subsets of 5-element set {1, 2, 3, 4, 5} is 2^5 (note that this includes an empty set as well as the original set {1, 2, 3, 4, 5}). Now, all subsets of {1, 2, 3, 4, 5} are the subsets of {0, 1, 2, 3, 4, 5} and does not include 0.

Does this make sense?
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Fresh Meat!!! [#permalink] New post 17 May 2014, 07:20
Bunuel wrote:
NGGMAT wrote:
Bunuel wrote:
3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

A. 16
B. 27
C. 31
D. 32
E. 64

Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.

Answer: D.


Dear Bunnel

I didnt understand this.

y didnt we take the set {1,2,3,4,5} into consideration and solve like the above qs? where did we get 2^5 from?


What do you mean by the red part?

As for 2^5: the number of subsets of n-element set is 2^n, thus the number of subsets of 5-element set {1, 2, 3, 4, 5} is 2^5 (note that this includes an empty set as well as the original set {1, 2, 3, 4, 5}). Now, all subsets of {1, 2, 3, 4, 5} are the subsets of {0, 1, 2, 3, 4, 5} and does not include 0.

Does this make sense?


By red part i meant that y we havent solved it like we did the below qs:

2. Set S contains 7 different letters. How many subsets of set S, including an empty set, contain at most 3 letters?

A. 29
B. 56
C. 57
D. 63
E. 64

1 empty set;
C^1_7=7 sets with one element;
C^2_7=21 sets with two elements;
C^3_7=35 sets with three element.

Total 1+7+21+35=64 sets


y did 2^n come into qs 3 and not qs 2?
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Re: Fresh Meat!!! [#permalink] New post 17 May 2014, 07:39
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NGGMAT wrote:
Bunuel wrote:
NGGMAT wrote:
Dear Bunnel

I didnt understand this.

y didnt we take the set {1,2,3,4,5} into consideration and solve like the above qs? where did we get 2^5 from?


What do you mean by the red part?

As for 2^5: the number of subsets of n-element set is 2^n, thus the number of subsets of 5-element set {1, 2, 3, 4, 5} is 2^5 (note that this includes an empty set as well as the original set {1, 2, 3, 4, 5}). Now, all subsets of {1, 2, 3, 4, 5} are the subsets of {0, 1, 2, 3, 4, 5} and does not include 0.

Does this make sense?


By red part i meant that y we havent solved it like we did the below qs:

2. Set S contains 7 different letters. How many subsets of set S, including an empty set, contain at most 3 letters?

A. 29
B. 56
C. 57
D. 63
E. 64

1 empty set;
C^1_7=7 sets with one element;
C^2_7=21 sets with two elements;
C^3_7=35 sets with three element.

Total 1+7+21+35=64 sets


y did 2^n come into qs 3 and not qs 2?


We could use 2^n for the second question too:

{The number of subsets with 0, 1, 2, or 3 terms} = {The total # of subsets} - {Subsets with 4, 5, 6, or 7 elements} = 2^7 - (C^4_7+C^5_7+C^6_7+C^7_7)=128-(35+21+7+1)=64.

But as you can see this approach is longer than the one used in my solution for that question.
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Re: Fresh Meat!!! [#permalink] New post 25 May 2014, 10:41
Hi Bunuel,

1. I went about the combinatorial approach and got 31 and saw your response below that states that one subset is the null set (empty set)

2. Now I also came across M16-23 in the GMAT club tests that states that "If the mean of the set S does not exceed mean of any subset of set S, which of the following must be true about set S?" And the right answer to that question is "all elements in set S are equal" and "the median of set S equals the mean of set S".

Aren't 1 and 2 contradictory? The only way in question M16-23 set S can have a mean more than mean of every subset including null set is if set S is null itself?

I am sure I am overthinking this and just need my caffeine.

Thanks,
Meera



Bunuel wrote:
jacg20 wrote:
Bunuel wrote:
3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

A. 16
B. 27
C. 31
D. 32
E. 64

Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.

Answer: D.



Hi Bunuel,

I did this exercise as follows:

I eliminate the 0, so i have the following set: (1,2,3,4,5). Now, i use combinatorics.

Set containing 5 elements: 5C5=1
Set containing 4 elements: 4C5=5
Set containing 3 elements: 3C5=10
Set containing 2 elements: 2C5=10
Set containing 1 elements: 1C5=5

So, the total of posibilites are 31. What am I missing here¿??

Thanks in advance


You are missing 1 empty set, which is a subset of the original set and also does not contain 0.

Hope it's clear.
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Re: Fresh Meat!!! [#permalink] New post 28 May 2014, 02:53
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Miraarun12345 wrote:
Hi Bunuel,

1. I went about the combinatorial approach and got 31 and saw your response below that states that one subset is the null set (empty set)

2. Now I also came across M16-23 in the GMAT club tests that states that "If the mean of the set S does not exceed mean of any subset of set S, which of the following must be true about set S?" And the right answer to that question is "all elements in set S are equal" and "the median of set S equals the mean of set S".

Aren't 1 and 2 contradictory? The only way in question M16-23 set S can have a mean more than mean of every subset including null set is if set S is null itself?

I am sure I am overthinking this and just need my caffeine.

Thanks,
Meera



Bunuel wrote:
3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

A. 16
B. 27
C. 31
D. 32
E. 64

Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.

Answer: D.



The point is that an empty set has no mean or the median, so when considering the subsets of S, we can ignore an empty set. Anyway this is out of the scope of the GMAT, so I wouldn't worry about it at all.
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Re: Fresh Meat!!! [#permalink] New post 20 Aug 2014, 00:35
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If x = 31, then f(31) = 5 and g(31) = 10: f(x) + g(x) = 5 + 10 = 15.

Why g(31) = 10 is not a prime number.

Bunuel wrote:
4. The functions f and g are defined for all the positive integers n by the following rule: f(n) is the number of positive perfect squares less than n and g(n) is the number of primes numbers less than n. If f(x) + g(x) = 16, then x is in the range:

A. 30 < x < 36
B. 30 < x < 37
C. 31 < x < 37
D. 31 < x < 38
E. 32 < x < 38

Perfect squares: 1, 4, 9, 16, 25, 36, ..,
Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, ...

If x = 31, then f(31) = 5 and g(31) = 10: f(x) + g(x) = 5 + 10 = 15.
If x = 32, then f(32) = 5 and g(32) = 11: f(x) + g(x) = 5 + 11 = 16.
...
If x = 36, then f(36) = 5 and g(36) = 11: f(x) + g(x) = 5 + 11 = 16.
If x = 37, then f(37) = 6 and g(37) = 11: f(x) + g(x) = 6 + 11 = 17.

Thus x could be 32, 33, 34, 35 or 36: 31<x<37.

Answer: C.

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Re: Fresh Meat!!! [#permalink] New post 20 Aug 2014, 02:31
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arindamsur wrote:
If x = 31, then f(31) = 5 and g(31) = 10: f(x) + g(x) = 5 + 10 = 15.

Why g(31) = 10 is not a prime number.

Bunuel wrote:
4. The functions f and g are defined for all the positive integers n by the following rule: f(n) is the number of positive perfect squares less than n and g(n) is the number of primes numbers less than n. If f(x) + g(x) = 16, then x is in the range:

A. 30 < x < 36
B. 30 < x < 37
C. 31 < x < 37
D. 31 < x < 38
E. 32 < x < 38

Perfect squares: 1, 4, 9, 16, 25, 36, ..,
Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, ...

If x = 31, then f(31) = 5 and g(31) = 10: f(x) + g(x) = 5 + 10 = 15.
If x = 32, then f(32) = 5 and g(32) = 11: f(x) + g(x) = 5 + 11 = 16.
...
If x = 36, then f(36) = 5 and g(36) = 11: f(x) + g(x) = 5 + 11 = 16.
If x = 37, then f(37) = 6 and g(37) = 11: f(x) + g(x) = 6 + 11 = 17.

Thus x could be 32, 33, 34, 35 or 36: 31<x<37.

Answer: C.


Why should it be?

g(n) is the number of primes numbers less than n: the number of prime numbers less than 31 is 10: 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29.
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Re: Fresh Meat!!! [#permalink] New post 20 Aug 2014, 02:38
Thanks again

I miss that word in hurry. number of prime numbers. I was looking for primes numbers. Thanks a lot.

Bunuel wrote:
arindamsur wrote:
If x = 31, then f(31) = 5 and g(31) = 10: f(x) + g(x) = 5 + 10 = 15.

Why g(31) = 10 is not a prime number.

Bunuel wrote:
4. The functions f and g are defined for all the positive integers n by the following rule: f(n) is the number of positive perfect squares less than n and g(n) is the number of primes numbers less than n. If f(x) + g(x) = 16, then x is in the range:

A. 30 < x < 36
B. 30 < x < 37
C. 31 < x < 37
D. 31 < x < 38
E. 32 < x < 38

Perfect squares: 1, 4, 9, 16, 25, 36, ..,
Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, ...

If x = 31, then f(31) = 5 and g(31) = 10: f(x) + g(x) = 5 + 10 = 15.
If x = 32, then f(32) = 5 and g(32) = 11: f(x) + g(x) = 5 + 11 = 16.
...
If x = 36, then f(36) = 5 and g(36) = 11: f(x) + g(x) = 5 + 11 = 16.
If x = 37, then f(37) = 6 and g(37) = 11: f(x) + g(x) = 6 + 11 = 17.

Thus x could be 32, 33, 34, 35 or 36: 31<x<37.

Answer: C.


Why should it be?

g(n) is the number of primes numbers less than n: the number of prime numbers less than 31 is 10: 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29.

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Re: Fresh Meat!!! [#permalink] New post 09 Sep 2014, 04:31
8. The product of a positive integer x and 377,910 is divisible by 3,300, then the least value of x is:

A. 10
B. 11
C. 55
D. 110
E. 330


Answer:D...


(x * 377910 ) / 3300... try cancel out common 377910's factor with 3310's factor.. whatever value left out in divisor should be X..

377910 - (3)*3*4199*(10)
3310- (3)*11*10*(10)..


u shall cancel the values 3 & 10 as both are available.. the left out in divisor is 11*10 then it should be the value of X. since 3310 is a divisor of the product
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Re: Fresh Meat!!! [#permalink] New post 09 Sep 2014, 04:36
Bunuel wrote:
2013gmat wrote:
Bunuel wrote:
6. If the least common multiple of a positive integer x, 4^3 and 6^5 is 6^6. Then x can take how many values?

A. 1
B. 6
C. 7
D. 30
E. 36

We are given that 6^6=2^{6}*3^{6} is the least common multiple of the following three numbers:

x;
4^3=2^6;
6^5 = 2^{5}*3^5;

First notice that x cannot have any other primes other than 2 or/and 3, because LCM contains only these primes.

Now, since the power of 3 in LCM is higher than the powers of 3 in either the second number or in the third, than x must have 3^{6} as its multiple (else how 3^{6} would appear in LCM?).

Next, x can have 2 as its prime in ANY power ranging from 0 to 6, inclusive (it cannot have higher power of 2 since LCM limits the power of 2 to 6).

Thus, x could take total of 7 values.

Answer: C.


Hi Bunuel,
x can take factor of 2 with power from 2 to 6 or no factor of 2. So the answer can be 6 too.
Please explain !
thanks


I don;t understand what you mean...

x can take the following 7 values:
3^6;
2*3^6;
2^2*3^6;
2^3*3^6;
2^4*3^6;
2^5*3^6;
2^6*3^6.



I am sure this question is quite easier than other question but i got some doubt before seeing this post. now clear.

Thanks bunuel
Re: Fresh Meat!!!   [#permalink] 09 Sep 2014, 04:36
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