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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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22 Jul 2016, 04:00
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56% (01:46) correct 44% (01:59) wrong based on 142 sessions

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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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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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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

E
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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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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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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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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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23 Jul 2016, 06:07
1
1
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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18 Sep 2017, 05:48
1
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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06 Feb 2018, 04:31
1
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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