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31 Aug 2013, 10:55
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: \(3^6\); \(2*3^6\); \(2^2*3^6\); \(2^3*3^6\); \(2^4*3^6\); \(2^5*3^6\); \(2^6*3^6\).
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31 Aug 2013, 22:30
\(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}=\frac{333}{999} + \frac{111}{999} + \frac{37}{999} + \frac{27}{999}=\frac{508}{999}=0.508508...\).
How do we get these fractions with a common denominator?



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01 Sep 2013, 12:08



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02 Sep 2013, 20:28
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. 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. Thanks in Advance, Rrsnathan



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03 Sep 2013, 02:26
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.
Answer: E. 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. Thanks in Advance, 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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18 Sep 2013, 02:57
Bunuel wrote: 25*1=25 and 25*13=325; 25*3=75 and 25*11=275; 25*5=125 and 25*9=225.
Answer: C. Hello, what about 175, 175 ? The question doesn't state they are distinct positive numbers.



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30 Sep 2013, 04:14
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. 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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30 Sep 2013, 05:41
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.
Answer: D. 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: howmanysubordinatesdoesmarciahave57169.html#p692676Hope it helps.
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12 Jan 2014, 22:15
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. Great questions Bunuel! Is there a trick to convert 1/37 > 27/999?
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13 Jan 2014, 00:17
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.
Answer: A. 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: mathnumbertheory88376.html (check Converting Fractions chapter).
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27 Feb 2014, 07:22
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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27 Feb 2014, 07:35
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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03 Apr 2014, 08:25
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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03 Apr 2014, 08:44
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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09 Apr 2014, 01:53
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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09 Apr 2014, 03:05
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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17 May 2014, 07:44
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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17 May 2014, 07:51
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 nelement set is 2^n, thus the number of subsets of 5element 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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17 May 2014, 08:20
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 nelement set is 2^n, thus the number of subsets of 5element 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 setsy did 2^n come into qs 3 and not qs 2?
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