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KindRope
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If m and n are positive integers such that m^3 is a factor of 3240 and 21 Sep 2026, 11:06
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65% (hard)

Question Stats:

42% (02:39) correct 58% (02:34) wrong based on 33 sessions

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If m and n are positive integers such that m^3 is a factor of 3240 and n^4 is a multiple of 108, what is the minimum possible value of the product mn ?


(A) 6
(B) 12
(C) 18
(D) 36
(E) 72
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A
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If m and n are positive integers such that m^3 is a factor of 3240 and 21 Sep 2026, 11:31
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KindRope
if m and n are positive integers such that m^3 is a factor of 3240 and n^4 is a multiple of 108, what is the minimum possible value of the product m.n ?


(A) 6
(B) 12
(C) 18
(D) 36
(E) 72
Show more
Step 1: Prime Factorization

3240 = 2^3 x 3^4 x 5^1
108 = 2^2 x 3^3

Step 2: Analyzing m^3 (factor of 3240)
[ltr][/ltr]
[ltr]Since m3m cubed, its prime powers cannot exceed the powers in 3240. Also, since m3 is a perfect cube, its powers must be multiples of 3 (i.e., 0, 3, 6...).

[*]Max possible power of 2 in m3 ≤ 3 --> 2^3
Max possible power of 3 in m3 ≤ 4 --> 3^3
Max possible power of 5 in m3 ≤ 1 --> 5^0 = 1
So, the maximum possible value for m3 = 23 x 33 = 63 --> m ≤ 6
(Since we want to minimize m.n., we should check what values m can take. m can be 1, 2, 3, or 6).
[*]Step 3: Analyzing n^4 (multiple of 108)
[*]
Since n^4 is a multiple of 108, its prime powers must be greater than or equal to the powers in
108(2^2 x 3^3) Since n^4 is a perfect 4th power, its powers must be multiples of 4 (i.e., 0, 4, 8...).

Min power of 2 in n^4 ≥ 2 --> 2^4
[ltr]Min power of 3 in n^4 ≥ 3 --> 3^4
So, the minimum possible value for n^4 = 2^4 x 3^4 = 6^4 --> Min n = 6[/ltr]




[/ltr]
[*]Step 4: Minimizing the Product m.n

We know the minimum possible value of n is fixed at 6
to minimize m.n, we need the smallest positive integer value for m.
Since m is a positive integer, its absolute minimum value is 1 (and 1^3 = 1, which is a factor of 3240)
therefore Min(m.n) = 1 x 6 = 6

Answer (A)
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If m and n are positive integers such that m^3 is a factor of 3240 and 21 Sep 2026, 11:34
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3240 = 3^4 * 2^3 * 5

M^3 is a factor which means lowest possible M is 1

Now 108=. 3^3* 2^2
N4 is a multiple of 108, to make 108* Some number= N^4, We can multiply 108 by 3 and 2, we get 3^4 * 2^4 , This is smallest N^4 we can arrive. N here is 6

MN=1*6=6
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If m and n are positive integers such that m^3 is a factor of 3240 and 21 Sep 2026, 15:53
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KindRope
If m and n are positive integers such that m^3 is a factor of 3240 and n^4 is a multiple of 108, what is the minimum possible value of the product mn ?


(A) 6
(B) 12
(C) 18
(D) 36
(E) 72
Show more
You can save yourself some work here by recognizing that the smallest positive factor of any integer is 1.

Since \(1^3 = 1\), 1 is the smallest possible value of m. No need to break 3240 into factors.

To find the smallest multiple of 108, you do need to prime factorize 108, which becomes \(2^2 * 3^3\). Since you're looking for \(n^4\), any primes that are in n will be repeated 4 times in \(n^4\). Thus, you only need one 2 and one 3 in n to "cover" the two 2's and three 3's in 108:
\(n=2*3=6\)

The smallest possible value of mn is thus \(1*6 = 6\). The answer is A.
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