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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

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.

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

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

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 post received KUDOS

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