Bunuel wrote:
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:
https://gmatclub.com/forum/does-the-deci ... 89566.htmlhttps://gmatclub.com/forum/any-decimal-t ... 01964.htmlhttps://gmatclub.com/forum/if-a-b-c-d-an ... 25789.htmlhttps://gmatclub.com/forum/700-question-94641.htmlhttps://gmatclub.com/forum/is-r-s2-is-a- ... 91360.htmlhttps://gmatclub.com/forum/pl-explain-89566.htmlhttps://gmatclub.com/forum/which-of-the- ... 88937.htmlHope it helps.
hi man
since 0<x<1, say x is a proper fraction ....
taking 2 statements together,
x cannot be 1/3, as 3 can divide 24, but cannot divide 28...
in the same line of reasoning, x cannot be 1/7, as 7 can divide 28, but cannot divide 24...
So, for both of the two statements to hold true, x cannot be 3 and/or 7, thus we are left with only 2s in the denominator. Sufficient ....
please say to me whether the reasoning is okay....
thanks in advance, man ...