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Does the decimal equivalent of m^3/n^2, where m and n are positive int

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Does the decimal equivalent of m^3/n^2, where m and n are positive int  [#permalink]

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New post 22 Jul 2016, 04:00
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Does the decimal equivalent of m^3/n^2, where m and n are positive integers, contain a finite amount of non-zero digits?

(1) (2n - 2)(2n - 4) is not divisible by 6
(2) m = 10^x where x is a positive integer

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Re: Does the decimal equivalent of m^3/n^2, where m and n are positive int  [#permalink]

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New post 22 Jul 2016, 05:32
Bunuel wrote:
Does the decimal equivalent of m^3/n^2, where m and n are positive integers, contain a finite amount of non-zero digits?

(1) (2n - 2)(2n - 4) is not divisible by 6
(2) m = 10^x where x is a positive integer



Given to find the decimal equivalent of m^3/n^2.

Stat 1: (2n - 2)(2n - 4) is not divisible by 6...not sure about this.

Stat 2: m = 10^x where x is a positive integer - x can take any values from 1 to infinite.

We need to know the values of m and n to get the division. Even if we combine both stats we can't get the unique value of m.

IMO E is correct option.

Waiting for more responses.
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Re: Does the decimal equivalent of m^3/n^2, where m and n are positive int  [#permalink]

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New post 22 Jul 2016, 08:17
(1) (2n - 2)(2n - 4) is not divisible by 6
comes to (n-1)(n-2)
which means they are consecutive
but not divisible by 6
NS
(2) m = 10^x where x is a positive integer
ca be 1000/2 1000/3 diff answers

combining, we have 1000/(something not divisible by 6)
can be 5 yes
can be 7 no

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Re: Does the decimal equivalent of m^3/n^2, where m and n are positive int  [#permalink]

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New post 22 Jul 2016, 09:27
Imo C.

From 1, 4 (n-1)(n-2) is not divisible by 6.. meaning (n-1)(n-2) are not multiple of 3. So n is definitely multiple of 3, being consecutive.
From 2, numerator is multiple of 10.
From 1 and 2 : we have multiple of 10/ multiple of 3
It can't be terminating decimal. E.g. 1000/ 3 , 10000/81,
10000/12 .. etc.
So both together are sufficient and answer is No.
B.t.w. this is my understanding, experts can supply OA and explanation. :)


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Re: Does the decimal equivalent of m^3/n^2, where m and n are positive int  [#permalink]

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New post 22 Jul 2016, 10:43
Bunuel wrote:
Does the decimal equivalent of m^3/n^2, where m and n are positive integers, contain a finite amount of non-zero digits?

(1) (2n - 2)(2n - 4) is not divisible by 6
(2) m = 10^x where x is a positive integer


Statement 1 doesn't tell anything about m. But for N, it does tell that n is divisible by 3.

Reason : Every third consecutive integer is divisible by 3. ( Excluding 0 among these integers). -- Insufficient.

Statement 2: No info about n but m doesn't have 3 in it. -- Insufficient.

Combining : We know that N is divisible by 3 and M doesn't have 3 in it. SO, m^3/n^2 is not a terminating decimal.

Thus, Answer is C.
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Re: Does the decimal equivalent of m^3/n^2, where m and n are positive int  [#permalink]

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New post 23 Jul 2016, 03:46
1
IMO C

An expression can be termination if the denominator is multiple of 3 and numerator is not a multiple of three

We can re write the first statement as 4(n-1)(n-2) and this is not divisible by 6(i.e. 2 or 3), this is only possible when n-1 and n-2 is not divisible by three. There could only one case where n-1 and n-2 is not divisible by 3 when n itself is 3.
However we don't know the value of m, it can be a multiple of 3 or an even number. So statement 1 is Insuff

From 2 we know that only m is a multiple of 10. So statement 2 is also in suff.

Now combining both the statements we can see that since n is 3 and m is multiple of 10 then definitely m^3/n^2 will be non terminating
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Re: Does the decimal equivalent of m^3/n^2, where m and n are positive int  [#permalink]

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New post 23 Jul 2016, 06:07
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Bunuel wrote:
Does the decimal equivalent of m^3/n^2, where m and n are positive integers, contain a finite amount of non-zero digits?

(1) (2n - 2)(2n - 4) is not divisible by 6
(2) m = 10^x where x is a positive integer



My Take.

(1) (2n - 2)(2n - 4) is not divisible by 6

the possible values are n=3K where k = 1,2,3..
also no information on m. ---insufficient

(2) m = 10^x where x is a positive integer
since m is a positive integer , here x can take values 1,2,3,4.. thus insufficient.

combining both
we have powers of 2 and 5 in the numerator, while we have at least one power of 3 in the denominator
(2^x*5^x)/3K

now since, we do not have any way to cancel the 3 in the denominator the result therefore will always be infinite amount of non zero digits.

Thus a definite No

C
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Re: Does the decimal equivalent of m^3/n^2, where m and n are positive int  [#permalink]

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New post 18 Sep 2017, 05:48
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This question can be quickly solved by putting values.
1. Put n= 3 and it satisfies statement 1. but we dont know value of m. Not sufficient
2. m = 10^x , we dont know value of n. Not sufficient

1 + 2 ----> we get 10^3x/ 3^2
Suppose x =1, then we get 1000/9 which is a recurring decimal. Sufficient and answer is no

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Re: Does the decimal equivalent of m^3/n^2, where m and n are positive int  [#permalink]

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New post 06 Feb 2018, 04:31
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Bunuel wrote:
Does the decimal equivalent of \(\frac{m^3}{n^2}\), where m and n are positive integers, contain a finite amount of non-zero digits?

(1) (2n - 2)(2n - 4) is not divisible by 6
(2) m = 10^x where x is a positive integer


It should be C

Question basically asks if \(\frac{m^3}{n^2}\) has a terminating decimal component.

1-

\((2n - 2)(2n - 4) = 4(n-1)(n-2)\)

Due to the \(4\) in the expression, we know this expression is divisible by \(2\), one of the factors of the divisor \(6\). We are now left with the other factor of \(6\) i.e.
\(3\). \(n-1\) and \(n-2\) are preceding 2 integers of \(n\) and if they are not divisible by \(3\) then \(n\) must be divisible by \(3\) (every 3rd integer is divisible by 3). However we do not know if this 3 is canceled out by any \(3\) from the numerator component. Hence 1 is insufficient.

2-
\(m = 10^x\) so numerator from stem would simply become \((10^x)^3\). This does not tell us anything about the prime factors of denominator so we cannot determine if the decimal terminates.

1+2. We know there is a \(3\) in denominator that does not cancel out so the expression cannot be a terminating decimal.
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Re: Does the decimal equivalent of m^3/n^2, where m and n are positive int &nbs [#permalink] 06 Feb 2018, 04:31
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