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# Fresh Meat!!!

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Math Expert
Joined: 02 Sep 2009
Posts: 43312

Kudos [?]: 139296 [13], given: 12783

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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!!!
_________________

Kudos [?]: 139296 [13], given: 12783

Math Expert
Joined: 02 Sep 2009
Posts: 43312

Kudos [?]: 139296 [0], given: 12783

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29 Aug 2017, 20:34
mykbdj 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.

When we say subset, is it okay to include an empty set even if it's not explicitly stated? If we include the empty set, answer is 31, else 32.

As you can see from the solution, the answer to your question is YES: empty set is a subset of every set.
_________________

Kudos [?]: 139296 [0], given: 12783

Manager
Joined: 18 Mar 2015
Posts: 102

Kudos [?]: 4 [0], given: 113

Location: India
Schools: ISB '19
GMAT 1: 600 Q47 V26
GPA: 3.59

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30 Aug 2017, 04:14
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.

Hu bunuel , I understood the logic but how the total came to 7 for 3power 6 and 2 power any value between 2 to 6?

Kudos [?]: 4 [0], given: 113

Math Expert
Joined: 02 Sep 2009
Posts: 43312

Kudos [?]: 139296 [0], given: 12783

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30 Aug 2017, 04:17
Expert's post
1
This post was
BOOKMARKED
r19 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.

Hu bunuel , I understood the logic but how the total came to 7 for 3power 6 and 2 power any value between 2 to 6?

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

Kudos [?]: 139296 [0], given: 12783

Intern
Joined: 16 Jul 2016
Posts: 13

Kudos [?]: 2 [0], given: 43

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12 Sep 2017, 01:38
[quote="Bunuel"]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.

Why is the option b incorrect? If x is equal to 1 and its raised to the power 0 then it will be 1 only.

Kudos [?]: 2 [0], given: 43

Math Expert
Joined: 02 Sep 2009
Posts: 43312

Kudos [?]: 139296 [0], given: 12783

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12 Sep 2017, 01:42
manik919 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.

Why is the option b incorrect? If x is equal to 1 and its raised to the power 0 then it will be 1 only.

Notice that the question asks "which of the following must be true" not "which of the following could be true". While II COULD be true it's not necessarily true (not always true). For example, if x=-1 and y is any even number, then $$(-1)^{even}=1$$.
_________________

Kudos [?]: 139296 [0], given: 12783

Re: Fresh Meat!!!   [#permalink] 12 Sep 2017, 01:42

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