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

Perfect squares: 1, 4, 9, 16, 25, 36, ..,
Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, ...

If x = 31, then f(31) = 5 and g(31) = 10: f(x) + g(x) = 5 + 10 = 15.
If x = 32, then f(32) = 5 and g(32) = 11: f(x) + g(x) = 5 + 11 = 16.
...
If x = 36, then f(36) = 5 and g(36) = 11: f(x) + g(x) = 5 + 11 = 16.
If x = 37, then f(37) = 6 and g(37) = 11: f(x) + g(x) = 6 + 11 = 17.

Thus x could be 32, 33, 34, 35 or 36: 31<x<37.

Answer: C.

It is given "f(n) is the number of positive perfect squares less than n" then X should be greater than 36, with this answer comes to D. ??? please correct if I am wrong & please correct.

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

Notice that if x=-1 and y is any even number, then \((-1)^{even}=1\), thus none of the options must be true.

Answer: E.

Bunuel
If we consider only I
i.e x=1
1 to power anything will be one so shouldn't the OA be A ?

The question asks: which of the following MUST be true, not COULD be true. Now, ask yourself is x = 1 ALWAYS true? Doesn't the solution you quote, gives an example when x = 1 is NOT true?

Yeah I got that but what's wrong with C ?
X=1 than x^y=1
y=0 than x^y=1
there doesn't exsist any other case for which x^y is not equal to 1

I'll copy my solution here:

Notice that if x=-1 and y is any even number, then \((-1)^{even}=1\), thus none of the options must be true.

Please read carefully.

Bunuel
By III x=1 or y=0
since x= -1 so y=0
Isn't (-1)^0 = 1 ?
As 0 is even integer

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.

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.

What is the logic behind this?

It should be: 5C5 + 5C4 + 5C3 + 5C2 + 5C1 + 1 = 1 + 5 + 10 + 10 + 5 + 1 = 32.

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

Given: \(\frac{377,910 *x}{3,300}=integer\).

Factorize the divisor: \(3,300=2^2*3*5^2*11\).

Check 377,910 for divisibility by 2^2: 377,910 IS divisible by 2 and NOT divisible by 2^2=4 (since its last two digits, 10, is not divisible by 4). Thus x must have 2 as its factor (377,910 is divisible only by 2 so in order 377,910*x to be divisible by 2^2, x must have 2 as its factor);

Check 377,910 for divisibility by 3: 3+7+7+9+1+0=27, thus 377,910 IS divisible by 3.

Check 377,910 for divisibility by 5^2: 377,910 IS divisible by 5 and NOT divisible by 25 (in order a number to be divisible by 25 its last two digits must be 00, 25, 50, or 75, so 377,910 is NOT divisible by 25). Thus x must have 5 as its factor.

Check 377,910 for divisibility by 11: (7+9+0)-(3+7+1)=5, so 377,910 is NOT divisible by 11, thus x must have 11 as its factor.

Therefore the least value of x is \(2*5*11=110\).

Answer: D.

If x and y are NOT co-prime, the GCD of \(25x\) and \(25y\) will be more than 25, not 25 as given.

Why aren't we including (9,5), (11,3) and (13,1). Moreover how would this question be solved if we were told that the least common multiple of two numbers is 350??

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1. Don't we have these pairs? If we include say (9,5) along with (5,9) do we get different two numbers?
2. If it were not given that the sum is 350, then the answer will be - infinitely many.