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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

1/37 = 27/999= 0.027(recurring)
1/27 = 37/999 = 0.037(recurring)
1/9 = 0.111(recurring)
1/3 = 0.333(recurring)

The total = 0.508(recurring) Thus the next 2 digits after the 99th digit = 5 then 0.

A.

Q3:

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

Question 9:

The WONDERFUL thing is that 37*27 = 999, therefore, we can write the sum as:
1/3 + 1/9 + 1/27 + 1/37 =
333/999 + 111/999 + 37/999 + 27/999 =
508/999 =
0.508508508508508...

Note that the decimals repeat themselves in period of 3. Since 101 divided by 3 gives remainder 2, we're looking for the position #2 in the repeated set 508.

Therefore, the digit will be 0.

Answer A

Question 10
The answer would be none of these E

I. X=1 ; x could be any no. if y=o
II. Y=0; x could become 1 therefore negating this statement
III. x=1 or y=0; well x could very well be -1; so not necessary

1. The length of the diagonal of square S is 15x, the lengths of the diagonals of rhombus R are 11x and 9x, where x is a positive integer.

The area of square S is d*d/2=(15x)^2/=225x^2/2 and the area of rhombus R is diagonal1*diagonal 2/2=11x*9x/2=99x^2/2.

So, the difference between them is (225x^2-99x^2)/2=63x^2. Since x is an integer, the difference must be divisible by 63. Therefore, I is possible for x=1. II is not possible, since if 63x^2=126--->x^2=2, which is not possible for an integer x. III is possible for x=2.

The answer is D.

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 and so they do not have any common factor except 1.
15, 17, 33 (=3*11), and 39 (=3*13) are factors of 18! and none of those can be a factor of 18!+1.
So, only 19 can be a factor of 18!+1

Correct answer is C.

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

Numbers are: x, 2^6, and (2^5)*(3^5)
LCM of the numbers = (2^6)*(3^6)
As 3^6 is not part of second and third numbers, it must be part of x.
So, lowest and highest values of x can be 3^6 and 6^6 {=(3^6)*(2^6)}.
So the values that x can take are: (3^6)*(2^0), (3^6)*(2^1), (3^6)*(2^2), (3^6)*(2^3), (3^6)*(2^4), (3^6)*(2^5), and (3^6)*(2^6).

Correct answer is C.

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

350 = 25 * 14
To have the GCD of two numbers to be 25, we need to split 14 into two co-prime numbers. Such pairs of numbers are: (1,13), (3,11), and (5,9).

Correct answer is C.

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

1/3 = 0.333333……… (recurring 3)
1/9 = 0.111111……… (recurring 1)
1/27 = 0.037037…………. (recurring 037)
1/37 = 0.027027………….. (recurring 027)

1/3 + 1/9 + 1/27 + 1/37 =0.508508…………….. (recurring 508)
--> 8 will be in every 3rd position
--> 8 will be in 99th position
--> 0 will be in 101st position

Correct answer is A.

Q1) 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

Answer is E

S diagonal : R diagonal 1 : R diagonal 2 = 15:11:9

Let's assume x is the unknown multiplier and the ratio of diagonals are 15x:11x:9x

\(Area of S =\frac{15x}{\sqrt{2}}\)

\(Area of R = 1/2 * 11x * 9x\)

\(Area of S - Area of R = \frac{225x - 99x}{2} = 63x\)

And I, II, II are multiples of 63 when x = 1, 2, 3

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
My answer is D
Let the diagonals be 15x,11x and 9x.
Area of a Rhombus is .5*d1*d2
 .5*11x*9x=> 11*9*x2/2. = (99x^2)/2
 Area of square with diagonal 15x is (225x^2)/2
Difference in areas is 63x^2.
Difference can be 63, for X = 1 ; 126 for x=root2 and 252 for x=2.
However when x= root2, the diagonals are not integers. Hence only 2 values are possible. 63 and 252

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
My answer C
Squares less than 36 : 1,4,9,16,25 = 5 primes less than 36 : 2,3,5,7,11,13,17,19,23,29,31. =11 and sum =16.
Hence the max value N has to be less than 37 as N =38 will increase the number of primes by 1 and squares by 1. And the sum will be 18
Minimum values of N has to be 32, as any value less than 32 will decrease the number of primes by 1.and the sum will be 15

Hi Bunuel,

I did this exercise as follows:

I eliminate the 0, so i have the following set: (1,2,3,4,5). Now, i use combinatorics.

Set containing 5 elements: 5C5=1
Set containing 4 elements: 4C5=5
Set containing 3 elements: 3C5=10
Set containing 2 elements: 2C5=10
Set containing 1 elements: 1C5=5

So, the total of posibilites are 31. What am I missing here¿??

Thanks in advance

You are missing 1 empty set, which is a subset of the original set and also does not contain 0.

Hope it's clear.

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

How do we get these fractions with a common denominator?

\(\frac{1}{3} =\frac{1*333}{3*333}=\frac{27}{999}\).

\(\frac{1}{9} =\frac{1*111}{9*111}=\frac{27}{999}\).

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

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

Following link might help for this problem: math-number-theory-88376.html (check Converting Fractions chapter).

Hello, what about 175, 175 ? The question doesn't state they are distinct positive numbers.