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Fresh Meat!!!

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Fresh Meat!!! [#permalink] New post   17 Apr 2013, 06:11
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Fresh Meat: Collection of Tough and Tricky Questions!!!


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


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


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


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


5. Which of the following is a factor of 18!+1?

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


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


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


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


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


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

The solutions can be found below on this page.

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Fresh Meat!!! [#permalink] New post   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 always co-prime, meaning that they don't share any common factors other than 1. For example, 20 and 21 are consecutive integers, and the only common factor they share is 1.

Since we can factor out each of the numbers 15, 17, 33 (3*11), and 39 (3*13) from \(18!\), these numbers are indeed factors of \(18!\) but cannot be factors of \(18!+1\). Therefore, only 19 could be a factor of \(18!+1\).

Answer: C.

Answer: C
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Re: Fresh Meat!!! [#permalink] New post   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

If you have a fraction where the denominator is 9, 99, 999, etc., the decimal equivalent will have a repeating pattern in the decimal part. The repeating part is just the numerator of the fraction, with enough leading zeroes added to match the number of 9s in the denominator.

Examples:

• \(\frac{2}{9} = 0.2222...\)

• \(\frac{3}{99} = 0.030303...\)

• \(\frac{45}{999} = 0.045045045...\)

Thus, for any fraction \(\frac{n}{999...}\), the decimal representation will be \(0.nnn...\), with the number \(n\) (padded with leading zeroes if necessary) recurring as many times as there are 9s in the denominator.

Notice that each term in the provided sum can be expressed as a fraction with a certain number of 9s in the denominator:

\(\frac{1}{3} = \frac{3}{9} =0.333...\)

\(\frac{1}{9} = 0.111...\)

\(\frac{1}{27} =\frac{37}{999} =0.037...\)

\(\frac{1}{37} =\frac{27}{999} =0.027...\)

Hence, \(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}=0.333... + 0.111... + 0.037... + 0.027... = 0.508...\)

Alternatively, since the denominators of each fraction are factors of 999, we can express all fractions with 999 in the denominator:

\(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}=\)

\(=\frac{333}{999} + \frac{111}{999} + \frac{37}{999} + \frac{27}{999}=\)

\(=\frac{508}{999}=\)

\(=0.508508...\).

In this decimal representation, the repeating part is 508. Therefore, every third digit is 8 (such as the 3rd, 6th, 9th, ... digits). The 102nd digit will be 8, because 102 is a multiple of 3. Hence, the digit preceding it, which is the 101st digit, must be 0.


Answer: A
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Re: Fresh Meat!!! [#permalink] New post   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 the greatest common factor of \(25x\) and \(25y\) will be more than 25.

Next, we know that \(25x+25y=350\). Reducing by 25 gives \(x+y=14\). Now, since \(x\) and \(y\) don't share any common factor but 1, the possible pairs \((x, y)\) can only be (1, 13), (3, 11), or (5, 9) (all other pairs like (2, 12), (4, 10), (6, 8), and (7, 7) share a 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 \(25*9=225\).

Answer: C.
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Re: Fresh Meat!!! [#permalink] New post   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.

Answer: D.
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Re: Fresh Meat!!! [#permalink] New post   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 (LCM) of three numbers:

\(x\);

\(4^3=2^6\);

\(6^5 = 2^{5}*3^5\);

First notice that \(x\) cannot include any prime factors other than 2 and/or 3 as the LCM is exclusively composed of these primes.

Next, since the exponent of 3 in the LCM is greater than the exponent of 3 in the other two numbers, \(x\) must have \(3^6\) as a factor. Otherwise, the factor \(3^{6}\) wouldn't appear in the LCM.

Furthermore, \(x\) can include the prime number 2 raised to any power from 0 to 6, inclusive (the LCM limits the power of 2 in \(x\) to 6, so it cannot be any higher).

Therefore, \(x\) can take on a total of 7 distinct values, namely:

\(3^6\);

\(2*3^6\);

\(2^2*3^6\);

\(2^3*3^6\);

\(2^4*3^6\);

\(2^5*3^6\);

\(2^6*3^6\).


Answer: C
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Re: Fresh Meat!!! [#permalink] New post   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

\(x^y=1\) implies we have one of the following three cases:

1. The base is 1, because \(1^y=1\), for all \(y\);

2. The exponent is 0, because \(x^0=1\), for any nonzero \(x\);

3. The base is -1 and the exponent is even, because \((-1)^x=1\), for any even \(x\)

Notice that if we have case 3, so if \(x=-1\) and \(y\) is any even number, then \((-1)^{even}=1\), and in this case none of the options must be true.


Answer: E
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Re: Fresh Meat!!! [#permalink] New post   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

Positive 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, the possible values for \(x\) are 32, 33, 34, 35, or 36, which means \(x\) is in the range \(31 \lt x \lt 37\).


Answer: C
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Re: Fresh Meat!!! [#permalink] New post   21 Apr 2013, 21:53
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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 diagonals 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).

The area of the square is given by \(area_{square}=\frac{d_s^2}{2}\), and the area of the rhombus is given by \(area_{rhombus}=\frac{d_1*d_2}{2}\).

The difference between the areas 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 for the difference to be 126, \(x\) should be \(\sqrt{2}\), which is not possible since \(x\) must be an integer.


Answer: D
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Re: Fresh Meat!!! [#permalink] New post   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 \(2*5*11=110\).

Answer: D.
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Fresh Meat!!! [#permalink] New post   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

Subsets of S, with at most 3 letters, will be:

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 elements.

Total: 1 + 7 + 21 + 35 = 64 sets.

Answer: E.
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Re: Fresh Meat!!! [#permalink] New post   17 Apr 2013, 06:28
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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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Re: Fresh Meat!!! [#permalink] New post   17 Apr 2013, 06:31
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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] New post   17 Apr 2013, 06:34
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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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Re: Fresh Meat!!! [#permalink] New post   17 Apr 2013, 06:35
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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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Re: Fresh Meat!!! [#permalink] New post   17 Apr 2013, 06:47
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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 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

The no of primes less than 30 = 10 primes. Also,the number of perfect squares less than 30 = 1,4,9,16,25 = 5. Thus, for 31<x<37, the total sum is 16.

C.
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Re: Fresh Meat!!! [#permalink] New post   17 Apr 2013, 06:49
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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 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:

perfect squares = 1 4 9 16 25 36
prime numbers = 2 3 5 7 11 13 17 19 23 19 31 37

The first number that makes f(x) + g(x) = 16 is 32 and the last is 36

IMO C. 31 < x < 37
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Re: Fresh Meat!!! [#permalink] New post   Updated on: 18 Apr 2013, 02:11
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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.

Originally posted by mau5 on 17 Apr 2013, 06:55.
Last edited by mau5 on 18 Apr 2013, 02:11, edited 2 times in total.
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Re: Fresh Meat!!! [#permalink] New post   17 Apr 2013, 06:58
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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?

\(2^6, 2^5*3^5,2^6*3^6\) and \(x\)
\(x\) MUST have a \(3^6\) and can have any \(2^n\) with \(0\leq{n}\leq{6}\). So x can have \(7\) values

IMO C. 7
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Re: Fresh Meat!!! [#permalink] New post   17 Apr 2013, 07:01
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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

We know that 377910 is not divisible by 11. Also, 3300 = 3*11*5^2*2^2. Now, as 377910 is divisible by 30, we are left with 11,5,2.Thus, the least value of x = 11*5*2 = 110.

D.
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