Q6.

LCM of x, 2^6 and (2*3)^5 = (2*3)^6
therefore x must at least have a 3^6 in it. Plus it can have other powers of 2 ranging from 2^0 to 2^6
that makes 7 possible values for X
And C

Q8

x * 2 * (3^2) * 5 * 13 * 17 * 19 / (2^2 * 3 * 5^2 * 11)
after simplification we are left with:

x * 3 * 13 * 17 * 19 / 2 * 5 * 11

therefore x must at least be divisible by 110 (2*5*11)

Ans D

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

The Q itself says that x is not equal to zero. Therefore, x can be +ve or -ve both okay.
now, option 1 says...... x=1 it is false becoz it can't be must be true as x can be one but it can be some other number as well. for ex. this also satisfies the condition given. 2^0=1. therefore it must not be true.

Option second says ...... II. x=1 and y=0 , similar to the above one the given condition can be true but it is not
must to be true as 2^0=1 again satisfies the condition given. Therefore False for me again.
Option -III. it says either x=1 or y=0 now this option fulfills the criteria as if x =1 then no matter the power raise to it it will alwz remains 1 or if y will be equal to zero then no matter the base but the exp. remains in every case. Hence,
III must be true ...... Therefore, C.
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If you don’t make mistakes, you’re not working hard. And Now that’s a Huge mistake.

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First of all, before solving this problem, we must remember that two consecutive integers have a GCF of 1. That means 2 consecutive integers only share one factor that is 1. They never share any other factor. therefore, the factors of 18! can never be the factors of 18!+1. Okay!!!

Now, lets take a look at the answer choices.
1. 15, it is a factor of 18! as 18! contains both 5 & 3. Therefore, Eliminated.
2. 17, it is a factor of 18! as 18! contains 17. Therefore, Eliminated.
3. 19, it is not a factor of 18! as 18! can't contain 19 in it. Therefore, Chosen... Correct.
4. 33, it is a factor of 18! as 18! contains both 11 & 3. Therefore, Eliminated.
5. 39, it is a factor of 18! as 18! contains both 13 & 3. Therefore, Eliminated.
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If you don’t make mistakes, you’re not working hard. And Now that’s a Huge mistake.

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Here, it is given that the LCM of x, 4^3 & 6^5 is 6^6.
we can write is like this way. LCM (x,2^6, 2^5*3^5) is 2^6*3^6.
Tha means that the LCM has 2^6*3^6 & we have 2^6, 2^5*3^5 therefore x must contain 3^6. alongwith 0,1,2,3,4,5,6. Therefore, x can be 7 different values. Therefore, 7. ........................C.
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If you don’t make mistakes, you’re not working hard. And Now that’s a Huge mistake.

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It is given that F(n) = # of perfect Squares < n &
G(n) = # of primes less than n. & it is given that F(n)+G(n) = 16 so, the values in the two functions can be anything,
so to keep the options limited, lets look at the answer choices.
A. 30 < x < 36 .........
Perfect Squares < 30 = 1 4 9 16 25 i.e. 5.
Prime Numbers < 36 = 2 3 5 7 11 13 17 19 23 19 31 i.e. 11.
Therefore, this satisfies the condition given. Therefore,............. A.
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If you don’t make mistakes, you’re not working hard. And Now that’s a Huge mistake.

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Here it is given that the GCF of two integers is 25 or 5^2. That means that they both share 5^2 only.
Now we can depict the intergers as 25a & 25b sharing only 25 & a & b don't share anything b/w them. okay !!
Now, as given in the question that sum of the integers is 350. therefore, 25a+25b=350 or 25(a+b)=350
or a+b =14. Now we look for those values of a & b those don't share anything b/w them. So,
a+b= 3+11 or 1+13 or 5+9.........so Three values. Therefore, 3. C..........................
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If you don’t make mistakes, you’re not working hard. And Now that’s a Huge mistake.

Question 2:

Subset with 0 letters: 1
Subset with 1 letter: 7
Subset with 2 letters: C(7,2) = 21
Subset with 3 letters: C(7,3) = 7.6.5/6 = 35

Total: 1 + 7 + 21 + 35 = 64

Answer: E

Question 4:
Easier if you test numbers (considering that f(x) + g(x) is always crescent).

x = 30 -> f(30) = 5, g(30) = 10 -> f+g = 15
x = 31 -> f(31) = 5, g(31) = 10 -> f+g = 15
x = 32 -> f(32) = 5, g(32) = 11 -> f+g = 16
x = 33 -> f(33) = 5, g(33) = 11 -> f+g = 16
x = 34 -> f(34) = 5, g(34) = 11 -> f+g = 16
x = 35 -> f(35) = 5, g(35) = 11 -> f+g = 16
x = 36 -> f(36) = 5, g(36) = 11 -> f+g = 16
x = 37 -> f(37) = 6, g(37) = 11 -> f+g = 17

Therefore: 30 < x < 37

Answer: B

Question 6
LCM (x,2^6,3^5.2^5) = (2^6).(3^6)

x = 2^a. 3^b
b must be equal to 6 (since 3^6 is a factor of the LCM)
a can be equal to 0,1,2,3,4,5,6

Therefore: 7 possibilities

Answer: C