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Zarrolou

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17 Apr 2013, 07:04

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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?

$GMD = 5^2$ 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

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17 Apr 2013, 07:23

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

$18!$ 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

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17 Apr 2013, 07:38

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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?

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!

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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.

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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

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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

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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

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19 Apr 2013, 02:02

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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.

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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 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.

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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

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.

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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

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.

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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

1/3 = 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.

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19 Apr 2013, 15:04

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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

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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

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

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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
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

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Bunuel

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.

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

Bunuel

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jacg20

Bunuel

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

Hope it's clear.

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31 Aug 2013, 22:30

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

How do we get these fractions with a common denominator?

Bunuel

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ygdrasil24

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

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

$\frac{1}{27} =\frac{1*37}{27*37}=\frac{27}{999}$.

$\frac{1}{37} =\frac{1*27}{37*27}=\frac{27}{999}$.

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

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Bunuel

25*1=25 and 25*13=325;
25*3=75 and 25*11=275;
25*5=125 and 25*9=225.

Answer: C.

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

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