If G represents the number of multiples of 3 between 3^30 and 3^50, in

Hi

Number of multiples of 3 between\(3^{30}&3^{50}\) will be \([fraction]3^{50}-3^{30}/3}+1=3^{29}*(3^20-1) +1\)...
Result is MULTIPLE of 3+1.... So it will not be multiple of 3 or 9..
Both II and III are WRONG.
3^29*(3^20-1)+1 is odd*even+odd = odd... Thus I is correct..

A

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

Since 3^30 and 3^50 both are divisible by 3 . So the number of multiples between 3^30 and 3^50 =
\((3^{50}- 3^{30})/3 +1\\
= 3 ^{29}(3^{20} - 1) +1\)

\(I. since 3^{20} is odd, 3^{20} -1 is even, 3 ^{29}(3^{20} - 1) +1 is odd.\)
\(II. 3 ^{29}(3^{20} - 1) is divisible by 3 , so 3 ^{29}(3^{20} - 1) +1 is not divisible by 3.\)
\(III. 3 ^{29}(3^{20} - 1) is divisible by 9 , so 3 ^{29}(3^{20} - 1) +1 is not divisible by 9.\)

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