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Re: If 0 < x < 1, is it possible to write x as a terminating dec
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22 Sep 2019, 07:42
OFFICIAL EXPLANATION All terminating decimals share one property: the denominator of the decimal (written as a fraction) must contain only 2’s and 5’s as prime factors. For example, 0.25 written as a fraction is 1/4. The denominator contains only 2’s.
(1): INSUFFICIENT. For 24x to be an integer, the prime factors of 24 must completely cancel out the factors in the denominator of x. 24 = 23 × 3. There is no way to know which prime factors the denominator of x actually contains. If it contains a 3, the decimal will not terminate. If the denominator only contains 2’s, then it will terminate.
For example, if x is 1/2, 24x = 12, and the decimal terminates (0.5). If, however, x = 1/3, 24x = 8, but the decimal does not terminate (0.3333….). The statement is insufficient.
(2): INSUFFICIENT. For 28x to be an integer, the prime factors of 28 must completely cancel out the factors in the denominator of x. 28 = 22 × 7. There is no way to know which prime factors the denominator of x actually contains. If it contains a 7, the decimal will not terminate. If the denominator only contains 2’s, then it will terminate.
For example, if x is 1/2, 28x = 14, and the decimal terminates (0.5). If, however, x = 1/7, 28x = 4, but the decimal does not terminate (0.14285….). The statement is insufficient.
(1) & (2): SUFFICIENT. For both 24x to be an integer and 28x to be an integer, the denominator of x can only contain prime factors that both 24 and 28 share. The only prime factors that 24 and 28 share are 2 and 2(22). Therefore, the denominator of x must only contain 2’s, and the decimal must terminate.
The correct answer is C.
