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If n and m are positive integers and m is a factor of 6^2, what is the

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Bunuel
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If n and m are positive integers and m is a factor of 6^2, what is the [#permalink] New post   20 Nov 2019, 00:07
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If n and m are positive integers and m is a factor of 6^2, what is the greatest possible number of integers that can be equal to both 3n and 6^2/m ?

A. Zero
B. One
C. Three
D. Four
E. Six

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chetan2u
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Re: If n and m are positive integers and m is a factor of 6^2, what is the [#permalink] New post   20 Nov 2019, 11:28
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Bunuel wrote:
If n and m are positive integers and m is a factor of 6^2, what is the greatest possible number of integers that can be equal to both 3n and 6^2/m ?

A. Zero
B. One
C. Three
D. Four
E. Six

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Now m can be any of the factors 1, 2,3,4,6,9,12,18,36...
But 36/m =3n.....12=mn
So for all m as factors of 12, n is a positive integer.
So m can be 1,2,3,4,6,12 and corresponding values of n too are 12,6,4,3,2,1
Hence 6 values of 3n.

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Re: If n and m are positive integers and m is a factor of 6^2, what is the [#permalink] New post   22 Nov 2019, 09:03
Question implies to me that M is in the expnent? ^(m/2) or not?
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Re: If n and m are positive integers and m is a factor of 6^2, what is the [#permalink] New post   24 Nov 2019, 23:48
Bunuel wrote:
If n and m are positive integers and m is a factor of 6^2, what is the greatest possible number of integers that can be equal to both 3n and 6^2/m ?

A. Zero
B. One
C. Three
D. Four
E. Six

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6^2 has 9 factors( \(2^2*3^2\))
Since 3n is a multiple of 3, for 3n to be equal to 6^2/m, we have to count only those factors which comprise 3 or multiple of 3
out of 9 multiple 2^0,2^1, 2^2 have nothing to do with multiple of 3
so 6 factors
E:)
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Re: If n and m are positive integers and m is a factor of 6^2, what is the [#permalink] New post   07 Dec 2022, 05:00
Both n and m are positive integers, and m is a factor of 26. How many integers are equal to 3n and 2^6/m at most? That is, find the number of Q that satisfies the integer Q=3n=2^6/m. However, if 3n=2^6/m, n=2^6/3m. n is divisible, so 2^6 should be divisible by 3m. However, divisibility can be reduced. There is no element in the numerator 2^6 that can be reduced by 3 in the denominator. It cannot be reduced or divisible, so this n does not exist, and Q for which the equation holds does not exist, and the number is 0.

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Re: If n and m are positive integers and m is a factor of 6^2, what is the [#permalink] New post   07 Dec 2022, 08:55
Asked: If n and m are positive integers and m is a factor of 6^2, what is the greatest possible number of integers that can be equal to both 3n and 6^2/m ?

3n = 6^2/m
nm = 6^2/3 = 12
36 = 6^2 = 2^2*3^2
m = {1,2,3,4,6,9,12,18,36}
6^/m = 36/m = 3n = {36,18,12,9,6,3}

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Re: If n and m are positive integers and m is a factor of 6^2, what is the [#permalink]
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