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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

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

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 _________________

It is beyond a doubt that all our knowledge that begins with experience.

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.

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:

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.

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.

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?

\(GMD = 5^2\) the numbers are \(5^2k\) and \(5^2q\) where q and k do not share any factor \(25k+25q=350\) \(25(k+q)=350\) \(k+q=14\) The possible numbers that summed give us 14 are: 1+13, 2+12,... those however must have no factor in common and those pairs are:1+13,3+11,5+9

IMO C.3 _________________

It is beyond a doubt that all our knowledge that begins with experience.

8. The product of a positive integer x and 377,910 is divisible by 3,300, then the least value of x is:

\(3,300=3*11*2*5*2*5\) \(377,910=37791*2*5\) \(377,910*x/3300=\frac{37791*2*5*x}{3*11*2*5*2*5}\) Now x must contain 2*5, because 37791 is divisible by 3 x must not contain a 3, because 37791 is not divisible by 11 x must have it. \(x=2*5*11=110\)

IMO D. 110 _________________

It is beyond a doubt that all our knowledge that begins with experience.

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

From the given inequality, for any y and x=1, we would have x^y = 1. Also, for any x(and not equal to 0) and y = 0, we would again have the same inequality.
_________________

10. If x is not equal to 0 and x^y=1, then which of the following must be true?

I. x=1 False \(100^0=1\) II. x=1 and y=0 False \(2^0=1\) III. x=1 or y=0 True Infact there are two cases: every number with a 0 exponent equals 1, and 1 raised to any exp equals 1.

IMO C. III only _________________

It is beyond a doubt that all our knowledge that begins with experience.

\(18!\) and \(18!+1\) are consecutives so they do not share any factor (except 1). A,B,D and E are factors of \(18!\) (ie:\(33=3*11\) both contained in \(18!\))

IMO C.19 _________________

It is beyond a doubt that all our knowledge that begins with experience.

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!
_________________

It is beyond a doubt that all our knowledge that begins with experience.

With 6 different elements in a set, total number of subsets = 2^6 = 64 With 5 different elements in a set, total number of subsets = 2^5 = 32 Hence from set of 6, if we do not include 0 in any set, that would be equivalent to considering just 5 elements out of 6 sets and how many subsets can be obtained from 5 elements. that should be 2^5 = 32 sub sets. And D

lets try with the options. A. x between 30 and 36 f(n) = 5. No of squares = 5 (1,4,9,16 and 25) g(n) = 11 (Primes are 2,3,5,7,11,13,17,19,23,29,31) f(n) + g(n) = 16