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Does the decimal equivalent of m^3/n^2, where m and n are positive int
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22 Jul 2016, 03:00
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54% (02:20) correct 46% (02:33) wrong based on 143 sessions
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Re: Does the decimal equivalent of m^3/n^2, where m and n are positive int
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22 Jul 2016, 04:32
Bunuel wrote: Does the decimal equivalent of m^3/n^2, where m and n are positive integers, contain a finite amount of nonzero 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
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22 Jul 2016, 07:17
(1) (2n  2)(2n  4) is not divisible by 6 comes to (n1)(n2) 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
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22 Jul 2016, 08:27
Imo C. From 1, 4 (n1)(n2) is not divisible by 6.. meaning (n1)(n2) 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. Sent from my SMN910H using Tapatalk



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Re: Does the decimal equivalent of m^3/n^2, where m and n are positive int
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22 Jul 2016, 09:43
Bunuel wrote: Does the decimal equivalent of m^3/n^2, where m and n are positive integers, contain a finite amount of nonzero 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
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23 Jul 2016, 02:46
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(n1)(n2) and this is not divisible by 6(i.e. 2 or 3), this is only possible when n1 and n2 is not divisible by three. There could only one case where n1 and n2 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
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23 Jul 2016, 05:07
Bunuel wrote: Does the decimal equivalent of m^3/n^2, where m and n are positive integers, contain a finite amount of nonzero 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
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18 Sep 2017, 04:48
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 Dont forget to give Kudos



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Re: Does the decimal equivalent of m^3/n^2, where m and n are positive int
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06 Feb 2018, 03:31
Bunuel wrote: Does the decimal equivalent of \(\frac{m^3}{n^2}\), where m and n are positive integers, contain a finite amount of nonzero digits?
(1) (2n  2)(2n  4) is not divisible by 6 (2) m = 10^x where x is a positive integer It should be CQuestion basically asks if \(\frac{m^3}{n^2}\) has a terminating decimal component. 1 \((2n  2)(2n  4) = 4(n1)(n2)\) 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. \(3\). \(n1\) and \(n2\) 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
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