Any decimal that has only a finite number of nonzero digits

A fraction r/s will only be a terminating decimal ONLY if it is of the form \(Numerator/ 2^m 5^n\), where n and m are non-negative.
Statement 1 gives the range of numerators, of which we are not concerned at all. Insufficient
Statement 2 gives the value of denominator which is of the form \(2^2\). Hence the fraction has to be a terminating decimal.
+1B

Carcass rightly said, you are a machine Bunuel.

Question, I understand that a terminating decimal has to be of the form \(2^x5^x\) but four is only in the form of \(2^n\) to be a terminating decimal it can meet either of the requirements?

Thanks,
Hunter

any decimal that has only a finite number of nonzero digits is a terminating decimal. for example, 24, 0.82, and 5.096 are three terminating numbers. If r and s are positive integers and the ratio is r/s is expressed as a decimal, is r/s a terminating decimal?
1. 90<r< 100
2. s = 4B

There is a mistake in the text. It says that any decimal number has only a finite number of nonzero digits, but this is not true. For this to be true, It must say: any decimal number has only a finite number of digits.
Do you agree?

The reason it says "non zero digits" specifically is because theoretically every decimal has infinite trailing 0s at the right of the decimal after the last non zero digit.

15.6 = 15.6000000000000000...
15.0903 = 15.0903000000000000000000...
15 = 15.0000000000000000000...

But all of 15.6, 15.0903 and 15 are terminating.

This problem is testing us on our knowledge of terminating decimals.

When solving this problem, we should remember that there is a special property of fractions that allows their decimal equivalents to terminate. In its most-reduced form, any fraction with a denominator whose prime factorization contains only 2s, 5s, or both produces decimals that terminate. A denominator with any other prime factors produces decimals that do not terminate. So to determine whether r/s is expressed as a terminating decimal, we need to determine whether the prime factorization of s contains only 2s, 5s, or both.

Statement One Alone:

90 < r < 100

Since statement one does not provide any information about s, we cannot determine whether r/s is expressed as a terminating decimal. If r = 91 and s = 1, then r/s is a terminating decimal. On the other hand, if r = 91 and s = 3, then r/s = 30.3333… and thus, r/s is not a terminating decimal. Statement one alone is not sufficient. We can eliminate answer choices A and D.

Statement Two Alone:

s = 4

Since we know that s = 4, we know that the prime factorization of s (2^2) only contains 2’s. Thus, r/s is expressed as a terminating decimal.

Answer: B

\(r,s\,\, \geqslant 1\,\,\,{\text{ints}}\)

\(\frac{r}{s}\,\,\,\mathop = \limits^? \,\,\,\,{\text{terminating}}\)

\(\left( 1 \right)\,\,90 < r < 100\,\,\,\,\left\{ \begin{gathered}\\
\,\left( {r,s} \right) = \left( {95,5} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\left( {\frac{{95}}{5} = \operatorname{int} } \right) \hfill \\\\
\,\left( {r,s} \right) = \left( {91,3} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\left( {\frac{{91}}{3} = 30\frac{1}{3} = 30.333 \ldots } \right) \hfill \\ \\
\end{gathered} \right.\)

\(\left( 2 \right)\,\,s = 4\)

\(\left( * \right)\,\,\,\,{\text{r/s}}\,\,\,{\text{division}}\,\,{\text{algorithm}}:\,\,\,\left\{ \begin{gathered}\\
\,r = qs + R\,\,\,\mathop = \limits^{s\, = \,4} \,\,\,4q + R \hfill \\\\
\,q\,\,\operatorname{int} \,\,\,,\,\,\,\,0\,\,\, \leqslant \,\,\,R\,\,\operatorname{int} \,\,\, \leqslant \,\,3\,\,\,\,\left( { = s - 1} \right) \hfill \\ \\
\end{gathered} \right.\)

\(\frac{r}{s}\,\,\,\,\mathop = \limits^{\,\left( * \right)} \,\,\,\,\frac{{4q + R}}{4} = q + \frac{R}{4}\,\, = \,\,\operatorname{int} \,\, + \,\,\frac{R}{4}\,\,\,\,\,\,\,\)

\(\frac{R}{4} = \,\,\,\left\{ {\begin{array}{*{20}{c}}\\
{\,\,\frac{0}{4}} \\ \\
{\,\,\frac{1}{4}} \\ \\
{\,\,\frac{2}{4}} \\ \\
{\,\,\frac{3}{4}} \\
\end{array}} \right.\begin{array}{*{20}{c}}\\
{\,\,{\text{if}}\,\,\,R = 0\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\left( {\frac{r}{4} = \operatorname{int} } \right)} \\ \\
{\,\,{\text{if}}\,\,\,R = 1\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\left( {\frac{r}{4} = \operatorname{int} \,\, + \,\,0.25} \right)} \\ \\
{\,\,{\text{if}}\,\,\,R = 2\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\left( {\frac{r}{4} = \operatorname{int} \,\, + \,\,0.5} \right)} \\ \\
{\,\,{\text{if}}\,\,\,R = 3\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\left( {\frac{r}{4} = \operatorname{int} \,\, + \,\,0.75} \right)} \\
\end{array}\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Hi Bunuel,

Please clarify below one :

A denominator with any other prime factors apart from 2s, 3s produces decimals that do not terminate.

Is the above statement true in each case irrespective of numerator ?

Also I want to know If the denominator has prime factors as 2,3,5 For ex denominator is 30. Now irrespective of numerator will it be terminate as the denominator has prime factors 2 and 3 .

Thanks

If the denominator has only 2's or/and 5's the the fraction will terminate irrespective of the numerator (obviously provided the numerator is an integer).

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