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# Fresh Meat!!!

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Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132499 [13], given: 12323

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17 Apr 2013, 06:11
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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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Kudos [?]: 132499 [13], given: 12323

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Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132499 [3], given: 12323

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21 Apr 2013, 21:58
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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.

_________________

Kudos [?]: 132499 [3], given: 12323

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132499 [4], given: 12323

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21 Apr 2013, 22:07
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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.

_________________

Kudos [?]: 132499 [4], given: 12323

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132499 [4], given: 12323

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21 Apr 2013, 22:29
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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.

_________________

Kudos [?]: 132499 [4], given: 12323

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132499 [6], given: 12323

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21 Apr 2013, 22:38
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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.

_________________

Kudos [?]: 132499 [6], given: 12323

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132499 [5], given: 12323

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21 Apr 2013, 22:56
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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.

_________________

Kudos [?]: 132499 [5], given: 12323

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Joined: 02 Sep 2009
Posts: 42249

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21 Apr 2013, 23:04
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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

We are told that the greatest common factor of two integers is 25. So, these integers are $$25x$$ and $$25y$$, for some positive integers $$x$$ and $$y$$. Notice that $$x$$ and $$y$$ must not share any common factor but 1, because if they do, then GCF of $$25x$$ and $$25y$$ will be more that 25.

Next, we know that $$25x+25y=350$$ --> $$x+y=14$$ --> since $$x$$ and $$y$$ don't share any common factor but 1 then (x, y) can be only (1, 13), (3, 11) or (5, 9) (all other pairs (2, 12), (4, 10), (6, 8) and (7, 7) do share common factor greater than 1).

So, there are only three pairs of such numbers possible:
25*1=25 and 25*13=325;
25*3=75 and 25*11=275;
25*5=125 and 25*9=225.

_________________

Kudos [?]: 132499 [0], given: 12323

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132499 [0], given: 12323

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21 Apr 2013, 23:27
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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

Given: $$\frac{377,910 *x}{3,300}=integer$$.

Factorize the divisor: $$3,300=2^2*3*5^2*11$$.

Check 377,910 for divisibility by 2^2: 377,910 IS divisible by 2 and NOT divisible by 2^2=4 (since its last two digits, 10, is not divisible by 4). Thus x must have 2 as its factor (377,910 is divisible only by 2 so in order 377,910*x to be divisible by 2^2, x must have 2 as its factor);

Check 377,910 for divisibility by 3: 3+7+7+9+1+0=27, thus 377,910 IS divisible by 3.

Check 377,910 for divisibility by 5^2: 377,910 IS divisible by 5 and NOT divisible by 25 (in order a number to be divisible by 25 its last two digits must be 00, 25, 50, or 75, so 377,910 is NOT divisible by 25). Thus x must have 5 as its factor.

Check 377,910 for divisibility by 11: (7+9+0)-(3+7+1)=5, so 377,910 is NOT divisible by 11, thus x must have 11 as its factor.

Therefore the least value of x is $$2*5*11=110$$.

_________________

Kudos [?]: 132499 [0], given: 12323

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132499 [0], given: 12323

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21 Apr 2013, 23:45
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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

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

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

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Kudos [?]: 132499 [0], given: 12323

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Joined: 02 Sep 2009
Posts: 42249

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21 Apr 2013, 23:48
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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.

_________________

Kudos [?]: 132499 [3], given: 12323

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Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132499 [2], given: 12323

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22 Apr 2013, 00:07
2
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Links to the solutions are in the original post here: fresh-meat-151046-100.html#p1213230
_________________

Kudos [?]: 132499 [2], given: 12323

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132499 [0], given: 12323

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22 Apr 2013, 00:07
Kudos points given for each correct solution.

Note that I cannot award more than 5 Kudos to the same person per day, so those of you who have more than 5 correct solutions please PM me tomorrow the links for which I owe you kudos points.

Thank you.

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Kudos [?]: 132499 [0], given: 12323

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Joined: 07 Sep 2010
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25 Apr 2013, 19:42
Hi Bunuel,
Which of the following is a factor of 18!+1?

A. 15
B. 17
C. 19
D. 33
E. None of These

Then, would it be possible to come at a conclusion that 19 will be the factor of 18!+1.
In the original question, we came to the answer by eliminating other choices.

Thanks
H
_________________

+1 Kudos me, Help me unlocking GMAT Club Tests

Kudos [?]: 1051 [0], given: 136

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132499 [0], given: 12323

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26 Apr 2013, 00:43
imhimanshu wrote:
Hi Bunuel,
Which of the following is a factor of 18!+1?

A. 15
B. 17
C. 19
D. 33
E. None of These

Then, would it be possible to come at a conclusion that 19 will be the factor of 18!+1.
In the original question, we came to the answer by eliminating other choices.

Thanks
H

Yes, we could get the correct answer with Wilson's theorem, but you don't need it for the GMAT.
_________________

Kudos [?]: 132499 [0], given: 12323

Intern
Joined: 25 Jul 2012
Posts: 39

Kudos [?]: 25 [0], given: 15

Concentration: Organizational Behavior, General Management
GMAT 1: 610 Q47 V26
GMAT 2: 640 Q49 V27
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29 May 2013, 05:59
1 3 +1 9 +1 27 +1 37 =333 999 +111 999 +37 999 +27 999 =508 999 =0.508508... .

would you please explain how 1/37 is 27/999
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PLAN >>> EXECUTE >>> MEASURE

Kudos [?]: 25 [0], given: 15

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Posts: 42249

Kudos [?]: 132499 [1], given: 12323

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29 May 2013, 07:41
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prinkashar wrote:
1 3 +1 9 +1 27 +1 37 =333 999 +111 999 +37 999 +27 999 =508 999 =0.508508... .

would you please explain how 1/37 is 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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Kudos [?]: 132499 [1], given: 12323

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30 May 2013, 06:23
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:

selecting 0 or more from n number of things is nC0+nC1+nC2+...+nCn
=> here it is selecting at most 3 letters so from 0 to 3 it is 7C0+7C1+7C2+7C3 = 64

Ans: E

Kudos [?]: 14 [0], given: 27

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30 May 2013, 06:32
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:

Subset means selecting 0 or more from given set. since 0 should be excluded we have 5 numbers in set.

selecting 0 to n from given set is 2^n => 2^5= 32.

Ans: 32

Kudos [?]: 14 [0], given: 27

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30 May 2013, 09:09
imhimanshu wrote:
Hi Bunuel,
Which of the following is a factor of 18!+1?

A. 15
B. 17
C. 19
D. 33
E. None of These

Then, would it be possible to come at a conclusion that 19 will be the factor of 18!+1.
In the original question, we came to the answer by eliminating other choices.

Thanks
H

Hi himanshu.
According to Wilson's Theorem, if P is a prime no. then the remainder when (p-1)! is divided by p is (p-1)
Therefore, 18! on division by 19 will give 18 as a remainder. Now 18+1 is divisible by 19 therefore answer to your query is 19.
add kudos if this helped you

Kudos [?]: 9 [0], given: 0

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132499 [1], given: 12323

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30 May 2013, 09:13
1
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Expert's post
gultrage wrote:
imhimanshu wrote:
Hi Bunuel,
Which of the following is a factor of 18!+1?

A. 15
B. 17
C. 19
D. 33
E. None of These

Then, would it be possible to come at a conclusion that 19 will be the factor of 18!+1.
In the original question, we came to the answer by eliminating other choices.

Thanks
H

Hi himanshu.
According to Wilson's Theorem, if P is a prime no. then the remainder when (p-1)! is divided by p is (p-1)
Therefore, 18! on division by 19 will give 18 as a remainder. Now 18+1 is divisible by 19 therefore answer to your query is 19.
add kudos if this helped you

Check here: fresh-meat-151046-100.html#p1217213
_________________

Kudos [?]: 132499 [1], given: 12323

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21 Jun 2013, 21:56
Hi Bunuel,
this seemed like a great way to earn kudos points,
Would you be having more questionaires like this in future also? as it sorta helps to boost up the kudos for people who have recently joined the forum and want to make it in time to get to the gmatclub tests by earning kudos.
_________________

Kudos [?]: 18 [0], given: 10

Re: Fresh Meat!!!   [#permalink] 21 Jun 2013, 21:56

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