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If n and m are positive integers and m is a factor of 6^2
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18 Nov 2019, 08:36
18 Nov 2019, 08:36
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65% (02:21) correct
34% (02:24) wrong based on 26 sessions
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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 \(\frac{6^2}{m}\) ?
A. Zero
B. One
C. Three
D. Four
E. Six
Kudos for the right answer and solution.
A. Zero
B. One
C. Three
D. Four
E. Six
Kudos for the right answer and solution.
Re: If n and m are positive integers and m is a factor of 6^2
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19 Nov 2019, 10:07
19 Nov 2019, 10:07
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3n=6^2/m
3nm = 6^2
nm = 12
The possible combinations are:
4 & 3
6 & 2
12 & 1
Therefore, we have six positive integer
3nm = 6^2
nm = 12
The possible combinations are:
4 & 3
6 & 2
12 & 1
Therefore, we have six positive integer
General Discussion
Re: If n and m are positive integers and m is a factor of 6^2
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18 Nov 2019, 21:34
18 Nov 2019, 21:34
5
Carcass 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 \(\frac{6^2}{m}\) ?
A. Zero
B. One
C. Three
D. Four
E. Six
Kudos for the right answer and solution.
A. Zero
B. One
C. Three
D. Four
E. Six
Kudos for the right answer and solution.
m is a factor of 36.
The factors(m) of 36 are 1,2,3,4,6,9,12,18,36.
Now we need to find the numbers such that 36/m = 3n.
so we need to find multiples of 3 when 36 is divided by its factors.
36/1 = 36(3n)
36/2 = 18(3n)
36/3 = 12(3n)
36/4 = 9(3n)
36/6 = 6(3n)
36/9 = 4
36/12 = 3(3n)
36/18 = 2
36/36 = 1
We see that there are a total of 6 numbers possible.
OA - E
