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Hi Bunuel,

Even I thought there was some problem with the question. Good that you were able to find the mistake :-D

Thanks.
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Clearly n = 2 and n = 3 would result in smaller numbers for 6^n/324. You can strike them off.

Put n = 4, and divide 324 into further factors, we get 6*6*6*6/6*6*3*3. Hence, the answer is C.
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Hi,

Bunuel, so the problem stem changed to "2" instead of "n"?

thanks.
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I understand the part where "6^n=2^n*3^n=324*k=2^2*3^4*k "
however I don't understand how you get k = 4 in "--> n_{min}=4 (for k=2^2=4)."

I understand the other solutions, however I would like to understand this one better, please. Thank you.

Bunuel
chintzzz
I tried solving the following question and arrived at (a). Please confirm if my answer is correct or the official answer (c) is correct.

What is the smallest positive integer n for which 324 is a factor of 6 to the power of 2.

A.2
B.3
C.4
D.5
E.6

Guess should be 6 to the power of \(n\).

Given: if \(6^n=324*k\), then \(n_{min}=?\)

\(6^n=2^n*3^n=324*k=2^2*3^4*k\) --> \(n_{min}=4\) (for \(k=2^2=4\)).

Answer: C.
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I understand the part where "6^n=2^n*3^n=324*k=2^2*3^4*k "
however I don't understand how you get k = 4 in "--> n_{min}=4 (for k=2^2=4)."

I understand the other solutions, however I would like to understand this one better, please. Thank you.

Bunuel
chintzzz
I tried solving the following question and arrived at (a). Please confirm if my answer is correct or the official answer (c) is correct.

What is the smallest positive integer n for which 324 is a factor of 6 to the power of 2.

A.2
B.3
C.4
D.5
E.6

Guess should be 6 to the power of \(n\).

Given: if \(6^n=324*k\), then \(n_{min}=?\)

\(6^n=2^n*3^n=324*k=2^2*3^4*k\) --> \(n_{min}=4\) (for \(k=2^2=4\)).

Answer: C.

\(6^n=2^n*3^n\).

\(324=2^2*3^4\).

Now, 6^n to be a multiple of 324, the powers of its primes must be at least as large as powers of primes in 324. Thus the least value of n for which 6^n is a multiple of 324 is 4.

Does this make sense?
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Yes, I guess I thought there was a magic formula where K = (2^n * 3^n)/((2^2*3^4). I wanted to learn how you would solve that.

Thanks again.
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Bunuel
chintzzz
I tried solving the following question and arrived at (a). Please confirm if my answer is correct or the official answer (c) is correct.

What is the smallest positive integer n for which 324 is a factor of 6 to the power of 2.

A.2
B.3
C.4
D.5
E.6

Guess should be 6 to the power of \(n\).

Given: if \(6^n=324*k\), then \(n_{min}=?\)

\(6^n=2^n*3^n=324*k=2^2*3^4*k\) --> \(n_{min}=4\) (for \(k=2^2=4\)).

Answer: C.

Hi,

The question asks for minimum value of n^th power of 6. Because 324 = 2^2*3^2 i.e. 324 = 6^2*3^2. So shouldn't n = 2?

TO
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Hi All,

This question can be solved with a bit of Number Property knowledge and TESTing THE ANSWERS.

We're asked for the SMALLEST positive integer N for which 324 is a FACTOR of 6^N.

In other words, which of these answers is SMALLEST and makes 6^N/324 is an integer.

Since we're dealing with positives, and the answer choices are numbers, we can just "brute force" the answer choices until we find a match....

A: N = 2 6^2 = 36 36/324 is NOT an integer. Eliminate A.

B: N = 3 6^3 = 216 216/324 is NOT an integer. Eliminate B.

C: N = 4 6^4 = 1296 1296/324 might "look" tough, but consider the patterns.... "12" is divisibly by "3" (4 times) and "96" is divisibly by "24" (also 4 times), so 1296/324 = 4. This IS an integer, so C MUST be the answer.

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The simplest way for me to solve the question is:

Prime factorisation of 324 = 2^2x3^4
Prime factorisation of 6 = 2x3

We need to have at least four factors of 2 and 3 on the number 6^N in order to end with an integer.

Answer C.
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chintzzz
What is the smallest positive integer n for which 324 is a factor of 6^n?

A. 2
B. 3
C. 4
D. 5
E. 6

324 = 4*81 = 2^2*3^4

i.e. 6^n=2^n*3^n must have atleast 2^2*3^4

i.e. n Min must be 4

Answer: Option C
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chintzzz
What is the smallest positive integer n for which 324 is a factor of 6^n?

A. 2
B. 3
C. 4
D. 5
E. 6



factorize 324 ---> \(2^2*3^4\)

factorize 6 ---> \(2^1*3^1\)

\(\frac{2^1*3^1}{2^2*3^4}\)

now so that \(6^n\) is divisible by 324 we need to add at least as many powers as 324 has

i.e. \(2^1*3^1\) = \(2^2*3^4\) (must be equal)

--> \(6^n\) = \(2^1*2^1*3^1*3^3\) = \(2^2*3^4\) YAY :)
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chintzzz
What is the smallest positive integer n for which 324 is a factor of 6^n?

A. 2
B. 3
C. 4
D. 5
E. 6

6^3=216=2/3*324
6^4=6*2/3*324=1296
1296/324=4
C
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chintzzz
What is the smallest positive integer n for which 324 is a factor of 6^n?

A. 2
B. 3
C. 4
D. 5
E. 6

Asked: What is the smallest positive integer n for which 324 is a factor of 6^n?

324 = 2^2 * 3^4

n = 6^4

iMO C
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