We could use 2^n for the second question too:

{The number of subsets with 0, 1, 2, or 3 terms} = {The total # of subsets} - {Subsets with 4, 5, 6, or 7 elements} = \(2^7 - (C^4_7+C^5_7+C^6_7+C^7_7)=128-(35+21+7+1)=64\).

But as you can see this approach is longer than the one used in my solution for that question.
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Hi Bunuel,

1. I went about the combinatorial approach and got 31 and saw your response below that states that one subset is the null set (empty set)

2. Now I also came across M16-23 in the GMAT club tests that states that "If the mean of the set S does not exceed mean of any subset of set S, which of the following must be true about set S?" And the right answer to that question is "all elements in set S are equal" and "the median of set S equals the mean of set S".

Aren't 1 and 2 contradictory? The only way in question M16-23 set S can have a mean more than mean of every subset including null set is if set S is null itself?

I am sure I am overthinking this and just need my caffeine.

Thanks,
Meera

The point is that an empty set has no mean or the median, so when considering the subsets of S, we can ignore an empty set. Anyway this is out of the scope of the GMAT, so I wouldn't worry about it at all.
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If x = 31, then f(31) = 5 and g(31) = 10: f(x) + g(x) = 5 + 10 = 15.

Why g(31) = 10 is not a prime number.

Why should it be?

g(n) is the number of primes numbers less than n: the number of prime numbers less than 31 is 10: 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29.
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I used a choose method and got 31. With the binary method you used, there is a set which contains no numbers. This increases the number to 32. But the question doesn't say there can be a subset with no numbers because 0 is also not included... but I like the binary method & may use it on these kinds of problems & just subtract 1 if that is necessary so THANK YOU. Thank you

Empty set is a subset of all non-empty sets.
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I believe the answer is wrong or please explain.
My reasoning IF x=36, there are 6 perfect squares NOT 5 (1,2,3,4,5,,6), and the sum would be 17. Thus the answer A is correct.

There are 5 positive perfect squares less than 36: 1 = 1^1, 4 = 2^2, 9 = 3^2, 16 = 4^4, and 25 = 5^2.
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Because we are looking for positive perfect squares LESS than 36.
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Can we see that since 5 is the first digit it will be the 100th digit and count from there or is that pure luck on this problem?

Pattern starts with 5 and repeats in block of three: 508 508 508... Thus 5 will be 100th digit because 100 = {multiple of 3} + 1 and 101st digit will be next number in pattern - 0.
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Bunuel,
please correct me if I am wrong,
the 7 values included below?
1) 3
2) 2
3) 2^2
4) 2^3
5) 2^4
6) 2^5
7) 2^6

Thank you so much

x can take the following 7 values:
\(3^6\);
\(2*3^6\);
\(2^2*3^6\);
\(2^3*3^6\);
\(2^4*3^6\);
\(2^5*3^6\);
\(2^6*3^6\).
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If x and y are NOT co-prime, the GCD of \(25x\) and \(25y\) will be more than 25, not 25 as given.
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Hu bunuel , I understood the logic but how the total came to 7 for 3power 6 and 2 power any value between 2 to 6?

This is answered many times...

x can take the following 7 values:
\(3^6\);
\(2*3^6\);
\(2^2*3^6\);
\(2^3*3^6\);
\(2^4*3^6\);
\(2^5*3^6\);
\(2^6*3^6\).
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[quote="Bunuel"]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.

Why is the option b incorrect? If x is equal to 1 and its raised to the power 0 then it will be 1 only.

Notice that the question asks "which of the following must be true" not "which of the following could be true". While II COULD be true it's not necessarily true (not always true). For example, if x=-1 and y is any even number, then \((-1)^{even}=1\).
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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

Solution -

Question can be re-written in simpler forms as - (x*377910)/3300
And what we have to check here is - Can 3300 fully divide the numerator ?

Upon further simplification, it becomes - (x*12597)/110

Now prime factorize the denominator - 2*5*11
None of these prime factors can further divide 12597. Therefore 12597 should be multiplied by 110 (i.e., x should be 110), in order for the expression to get fully reduced.

Hence D is the answer.