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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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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
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
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
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
