Re: If 0 < x < 1, is it possible to write x as a terminating dec

If 0 < x < 1, is it possible to write x as a terminating decimal?

(1) 24x is an integer --> 24x=m, where m an integer --> x=\frac{m}{24}=\frac{m}{2^3*3}, If m is a multiple of 3, then the answer is YES, else the answer is NO. Not sufficient.

(2) 28x is an integer --> 28x=n, where n an integer --> x=\frac{n}{28}=\frac{n}{2^2*7}, If n is a multiple of 7, then the answer is YES, else the answer is NO. Not sufficient.

(1)+(2) x=\frac{m}{2^3*3}=\frac{n}{2^2*7} --> \frac{m}{n}=\frac{2*3}{7} --> m IS a multiple of 3 (as well as n IS multiple of 7). Sufficient.

Answer: C.

Theory:
Reduced fraction \frac{a}{b} (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only b (denominator) is of the form 2^n5^m, where m and n are non-negative integers. For example: \frac{7}{250} is a terminating decimal 0.028, as 250 (denominator) equals to 2*5^2. Fraction \frac{3}{30} is also a terminating decimal, as \frac{3}{30}=\frac{1}{10} and denominator 10=2*5.

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \frac{x}{2^n5^m}, (where x, n and m are integers) will always be the terminating decimal.

We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \frac{6}{15} has 3 as prime in denominator and we need to know if it can be reduced.

Questions testing this concept:
does-the-decimal-equivalent-of-p-q-where-p-and-q-are-89566.html
any-decimal-that-has-only-a-finite-number-of-nonzero-digits-101964.html
if-a-b-c-d-and-e-are-integers-and-p-2-a3-b-and-q-2-c3-d5-e-is-p-q-a-terminating-decimal-125789.html
700-question-94641.html
is-r-s2-is-a-terminating-decimal-91360.html
pl-explain-89566.html
which-of-the-following-fractions-88937.html

Hope it helps.