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Re: PS - Terminating Decimal [#permalink]
04 Feb 2005, 22:13

terminating decimal is the decimal number which terminates at a certain point. the fraction whose denominator is ends at 5 or in multiple of 5 are terminating decimals as compared with the fractions ending at integers other than 5 or its multiples.

the OA should be C, but E terminates earlier than C.

The fraction will terminate if and only if the denominator has for prime divisors only 2 and 5, that is, if and only if the denominator has the form 2^a * 5^b for some exponents a >= 0 and b >= 0. The number of decimal places until it terminates is the larger of a and b.

The proof of this lies in the fact that every terminating decimal has the form n/10^e, for some e >= 0 (e is the number of places to the right that the decimal point must be moved to give you an integer, and n is that integer), and every fraction of that form has a terminating decimal found by writing down n and moving the decimal point e places to the left. Now when you cancel common factors from n/10^e = n/(2*5)^e = n/(2^e*5^e), it may reduce the exponents in the denominator, but that is all that can happen.

Basically you turn 1/128 to (1/2)^7, you know that 1/2 is a terminating decimal, therefore (1/2)^7 must also be. The denominator of the others has factors of 3 or 7 and cannot be terminating decimals.

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