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

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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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.

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

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

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

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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