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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

Kudos points for each correct solution!!!
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21 Apr 2013, 21:38
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5. Which of the following is a factor of 18!+1?

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

18! and 18!+1 are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

Now, since we can factor out each 15, 17, 33=3*11, and 39=3*13 out of 18!, then 15, 17, 33 and 39 ARE factors of 18! and are NOT factors of 18!+1. Therefore only 19 could be a factor of 18!+1.

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21 Apr 2013, 22:48
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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.

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21 Apr 2013, 20:53
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SOLUTIONS:

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

Given that the ratio of the diagonal is $$d_s:d_1:d_2=15x:11x:9x$$, for some positive integer x (where $$d_s$$ is the diagonal of square S and $$d_1$$ and $$d_2$$ are the diagonals of rhombus R).

$$area_{square}=\frac{d^2}{2}$$ and $$area_{rhombus}=\frac{d_1*d_2}{2}$$.

The difference is $$area_{square}-area_{rhombus}=\frac{(15x)^2}{2}-\frac{11x*9x}{2}=63x^2$$.

If x=1, then the difference is 63;
If x=2, then the difference is 252;
In order the difference to be 126 x should be $$\sqrt{2}$$, which is not possible.

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21 Apr 2013, 21:07
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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.

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17 Apr 2013, 05:28
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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

Side square = 15x $$AreaS = \frac{15^2}{2}x^2$$
Diagonals= 9x, 11x$$AreaR = \frac{11*9*x^2}{2}$$
Difference = $$\frac{15^2x^2-11*9x^2}{2}= \frac{126x^2}{2}= 63x^2$$
$$63=3*3*7$$
if x=1 diff = 63 possible and easy to see
$$126=2*3*3*7$$ x sould be $$\sqrt{2}$$ => no integer
$$252=2*2*3*3*7$$ x=2 possible

IMO D. I and III only
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17 Apr 2013, 16:14
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Question 10
The answer would be none of these E

I. X=1 ; x could be any no. if y=o
II. Y=0; x could become 1 therefore negating this statement
III. x=1 or y=0; well x could very well be -1; so not necessary
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21 Apr 2013, 20:58
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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.

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21 Apr 2013, 21:29
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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.

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21 Apr 2013, 21:56
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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.

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17 Apr 2013, 05:34
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3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

It's like asking how many subsets has {1,2,3,4,5}
$$2^5=32$$

IMO D. 32
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17 Apr 2013, 05:35
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5. Which of the following is a factor of 18!+1?

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

18! and 18!+1 are consecutive integers, thus co-prime. All options apart from C are present in 18!. Thus 19 is the only factor present in 18!+1.

C.
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17 Apr 2013, 05:55
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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

The two numbers can be represented as 25a and 25b, where a and b are co-prime.Also, 25(a+b) = 350 --> (a+b) = 14
Thus, a=1,b=13 or a=3,b=11 or a=9,b=5.

C.
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Last edited by mau5 on 18 Apr 2013, 01:11, edited 2 times in total.
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17 Apr 2013, 06:38
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9. What is the 101st digit after the decimal point in the decimal representation of 1/3 + 1/9 + 1/27 + 1/37?

1/3=0.333
1/9=0.333/3=0.111
1/27=0.037 and then repeats
1/37=0.027 and then repeats
We can work on the first 3 digits: 0.111+0.333+0.027+0.037=0.508
After the 0 we have at first place a 5, second a 0, third an 8; and so on 4th=5, 5th=0, 6th=8. Every 10th position we have a "change" 10th=5 20th=0 30th=8 and so on
100th=5 and finally 101st=0

IMO A.0

Thanks for the set Bunuel!
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17 Apr 2013, 15:06
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Question 9:

The WONDERFUL thing is that 37*27 = 999, therefore, we can write the sum as:
1/3 + 1/9 + 1/27 + 1/37 =
333/999 + 111/999 + 37/999 + 27/999 =
508/999 =
0.508508508508508...

Note that the decimals repeat themselves in period of 3. Since 101 divided by 3 gives remainder 2, we're looking for the position #2 in the repeated set 508.

Therefore, the digit will be 0.

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19 Apr 2013, 09:20
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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

Numbers are: x, 2^6, and (2^5)*(3^5)
LCM of the numbers = (2^6)*(3^6)
As 3^6 is not part of second and third numbers, it must be part of x.
So, lowest and highest values of x can be 3^6 and 6^6 {=(3^6)*(2^6)}.
So the values that x can take are: (3^6)*(2^0), (3^6)*(2^1), (3^6)*(2^2), (3^6)*(2^3), (3^6)*(2^4), (3^6)*(2^5), and (3^6)*(2^6).

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19 Apr 2013, 09:21
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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

1/3 = 0.333333……… (recurring 3)
1/9 = 0.111111……… (recurring 1)
1/27 = 0.037037…………. (recurring 037)
1/37 = 0.027027………….. (recurring 027)

1/3 + 1/9 + 1/27 + 1/37 =0.508508…………….. (recurring 508)
--> 8 will be in every 3rd position
--> 8 will be in 99th position
--> 0 will be in 101st position

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21 Apr 2013, 23:07
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Links to the solutions are in the original post here: fresh-meat-151046-100.html#p1213230
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31 Aug 2013, 09:55
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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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17 Apr 2013, 05:31
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2. Set S contains 7 different letters. How many subsets of set S, including an empty set, contain at most 3 letters?

At most 3 letters = 0 letters or 1 letter or 2 letters or 3 letters
0=1
1=7C1=7
2=7C2=21
3=7C3=35
$$1+7+21+35=64$$

IMO E. 64
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Re: Fresh Meat!!!   [#permalink] 17 Apr 2013, 05:31

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