If G represents the number of multiples of 3 between 3^30 and 3^50, inclusive, then G must be:

I. Odd
II. Divisible by 3
III. Divisible by 9

A. I only
B. II only
C. I and II only
D. II and III only
E. I, II, and III
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I converted the problem to a simpler one to check what the outcome was if I used multiples of 3 between 3^1 and 3^5.

The number of multiples of 3 for the above is (3^5-3^1)/3 = (243-3)/3 = 80
Now, the question says between, so the number of multiples have to be subtracted by 1 as one of the extremes would be included = 80-1 = 79

So, for this case, only option that fits is I.
Hence, I deduced that for multiples between 3^30 and 3^50 it would be the same.

Waiting to see official answer, and some alternate methods to solve!

Let’s analyze each Roman numeral.

I. Odd

We can use the following formula:

(largest multiple of 3 in the set - smallest multiple of 3 in the set)/3 + 1 = number of multiples of 3

(3^50 - 3^30)/3 + 1

3^50/3 - 3^30/3 + 1

3^49 - 3^29 + 1

3^29(3^20 - 1) + 1

Since 3^29 is odd and 3^20 - 1 is even, 3^29(3^20 - 1) is even, since odd x even = even. So, 3^29(3^20 - 1) + 1 is odd, since 1 more than an even number is odd. Thus, G is an odd number. Roman numeral I is correct.

II. Divisible by 3

From the calculation in Roman numeral I, we see that G = 3^29(3^20 - 1) + 1 and we see that the first term, 3^29(3^20 - 1), is divisible by 3, but the second term, 1, isn’t. Thus, the sum is not divisible by 3. Roman numeral II is not correct.

II. Divisible by 9

From the analysis in Roman numeral II, we see that G is not divisible by 3. If a number is not divisible by 3, it is not divisible by 9. Thus, Roman numeral III is also not correct.

Answer: A
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