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Math Expert V
Joined: 02 Sep 2009
Posts: 59632

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2
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.

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.
thanks

I don;t understand what you mean...

x can take the following 7 values:
$$3^6$$;
$$2*3^6$$;
$$2^2*3^6$$;
$$2^3*3^6$$;
$$2^4*3^6$$;
$$2^5*3^6$$;
$$2^6*3^6$$.
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Posts: 87
Location: United States
Concentration: Entrepreneurship, Operations
Schools: ISB '15
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$$\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}=\frac{333}{999} + \frac{111}{999} + \frac{37}{999} + \frac{27}{999}=\frac{508}{999}=0.508508...$$.

How do we get these fractions with a common denominator?
Math Expert V
Joined: 02 Sep 2009
Posts: 59632

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ygdrasil24 wrote:
$$\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}=\frac{333}{999} + \frac{111}{999} + \frac{37}{999} + \frac{27}{999}=\frac{508}{999}=0.508508...$$.

How do we get these fractions with a common denominator?

$$\frac{1}{3} =\frac{1*333}{3*333}=\frac{27}{999}$$.

$$\frac{1}{9} =\frac{1*111}{9*111}=\frac{27}{999}$$.

$$\frac{1}{27} =\frac{1*37}{27*37}=\frac{27}{999}$$.

$$\frac{1}{37} =\frac{1*27}{37*27}=\frac{27}{999}$$.

Following link might help for this problem: math-number-theory-88376.html (check Converting Fractions chapter).
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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.

Hi Bunuel,

As per the question which of the following must be true.

So as per the given choice B) II only is true right
where 1^0 = 1 as X =1 and Y=0 given.

As ur explanation gives another chance as
X coud be = -1 , and Y = any even.

Please clarify where i am wrong.

Rrsnathan
Math Expert V
Joined: 02 Sep 2009
Posts: 59632

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rrsnathan 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.

Hi Bunuel,

As per the question which of the following must be true.

So as per the given choice B) II only is true right
where 1^0 = 1 as X =1 and Y=0 given.

As ur explanation gives another chance as
X coud be = -1 , and Y = any even.

Please clarify where i am wrong.

Rrsnathan

x=1 and y=0 indeed satisfies x^y=1, but the question asks "which of the following must be true". So, this option is NOT necessarily true, because x can be -1 and y any even number.
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Bunuel wrote:
25*1=25 and 25*13=325;
25*3=75 and 25*11=275;
25*5=125 and 25*9=225.

Hello, what about 175, 175 ? The question doesn't state they are distinct positive numbers.
Math Expert V
Joined: 02 Sep 2009
Posts: 59632

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1
muzammil wrote:
Bunuel wrote:
25*1=25 and 25*13=325;
25*3=75 and 25*11=275;
25*5=125 and 25*9=225.

Hello, what about 175, 175 ? The question doesn't state they are distinct positive numbers.

The point is that if the two integers are 175 and 175, then their greatest common divisor going to be 175, not 25 as given in the stem.

Does this make sense?
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Posts: 143
Concentration: Sustainability, Entrepreneurship
GMAT 1: 770 Q50 V44 WE: Business Development (Internet and New Media)

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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.

I would like to request some some help with this question..

Could you please elaborate on the theory behind 2^n?
Isnt the null set considered to be a subset of this set?
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Math Expert V
Joined: 02 Sep 2009
Posts: 59632

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Transcendentalist 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.

I would like to request some some help with this question..

Could you please elaborate on the theory behind 2^n?
Isnt the null set considered to be a subset of this set?

Consider simpler set: {a, b, c}. How many subsets does it have? Each of a, b, and c has two choices either to be included in subsets or not. Thus total of 2^38 subsets (including an empty set). The subsets are:

{a, b, c} --> each is included;
{a, b} --> a and b are included;
{a, c};
{b, c};
{a};
{b};
{c};
{} empty set, none is included.

2^3=8 subsets.

Harder question to practice about the same concept: how-many-subordinates-does-marcia-have-57169.html#p692676

Hope it helps.
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Posts: 166
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GMAT 1: 640 Q44 V34 GMAT 2: 710 Q48 V40 GMAT 3: 720 Q49 V40 GPA: 3.16
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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.

Great questions Bunuel!

Is there a trick to convert 1/37 --> 27/999?
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Math Expert V
Joined: 02 Sep 2009
Posts: 59632

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m3equals333 wrote:
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.

Great questions Bunuel!

Is there a trick to convert 1/37 --> 27/999?

$$\frac{1}{3} =\frac{1*333}{3*333}=\frac{27}{999}$$.

$$\frac{1}{9} =\frac{1*111}{9*111}=\frac{27}{999}$$.

$$\frac{1}{27} =\frac{1*37}{27*37}=\frac{27}{999}$$.

$$\frac{1}{37} =\frac{1*27}{37*27}=\frac{27}{999}$$.

Following link might help for this problem: math-number-theory-88376.html (check Converting Fractions chapter).
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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.

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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Math Expert V
Joined: 02 Sep 2009
Posts: 59632

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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.

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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1
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.

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?
Math Expert V
Joined: 02 Sep 2009
Posts: 59632

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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.

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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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.

Hi Bunuel,

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

Thanks..
Math Expert V
Joined: 02 Sep 2009
Posts: 59632

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1
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.

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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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.

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?
Math Expert V
Joined: 02 Sep 2009
Posts: 59632

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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.

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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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.

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? Re: Fresh Meat!!!   [#permalink] 17 May 2014, 08:20

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# Fresh Meat!!!  