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 co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

Now, since we can factor out each 15, 17, 33=3*11, and 39=3*13 out of 18!, then 15, 17, 33 and 39 ARE factors of 18! and are NOT factors of 18!+1. Therefore only 19 could be a factor of 18!+1.

Answer: 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?

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!

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

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

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

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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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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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

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.
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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

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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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

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

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.
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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 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

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

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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Q2.

# Empty Set = 1
# Sets with 1 element = 7C1 = 7
# Sets with 2 elements = 7C2 = 21
# Sets with 3 elements = 7C3 = 35

# Subsets containing at most 3 letters = 1 + 7 + 21 + 35 = 64
Ans E

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