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Any decimal that has only a finite number of nonzero digits [#permalink]

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17 Feb 2010, 16:58

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Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 36, 0.72, and 3.005 are terminating decimals. If a, b, c, d and e are non-negative integers and p=2^a*3^b and q=2^c*3^d*5^e, is p/q a terminating decimal?

Interesting question and what nitish suggested was a good formula but when you apply it, the answer should actually be different.

The answer should be A. Only statement 1 is sufficient to say that the ratio p/q is non-terminating definitively.

Stmt-1 deals with the relationship between a and c, so we know clearly that 2 will not be there in the denominator.

Stmt-2 on the other hand relates b with d, so we don't know if 2 will be there in the denominator or not.

2^a/2^c = 2^(a - c) when a > c in the numerator and will be 1/2^(c - a) when a < c.

Can you explain why A is the answer?

I guess A only tells us that 2 wont be there in the denominator but does not tell us anything about 3, and now if 3 will be there in the denominator it will be non terminating decimal but if 3 wont be there then it will be a terminating decimal and hence its not sufficient

on the other hand st 2 clearly tells that 3 wont be there in the denominator and hence its sufficient.

Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 36, 0.72, and 3.005 are terminating decimals. If a, b, c, d and e are non-negative integers and p = \(2^a3^b\) and q = \(2^c3^d5^e\), is p/q a terminating decimal?

(1) a > c

(2) b > d

IMO B,

For any number to be a terminating decimal. Denominator should be in the format 2^x * 5^y.

Nitish, Is this a rule for a number to be terminating decimal ?

I'm not Nitish, but if I can answer your question, then yes, it is a rule. For the number to be a terminating decimal in denominator it has to have 2^x*5^y and x or y can be 0

Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 36, 0.72, and 3.005 are terminating decimals. If a, b, c, d and e are non-negative integers and p = \(2^a3^b\) and q = \(2^c3^d5^e\), is p/q a terminating decimal?

(1) a > c

(2) b > d

IMO B,

For any number to be a terminating decimal. Denominator should be in the format 2^x * 5^y.

Nitish, Is this a rule for a number to be terminating decimal ?

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.

Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 36, 0.72, and 3.005 are terminating decimals. If a, b, c, d and e are non-negative integers and p = \(2^a3^b\) and q = \(2^c3^d5^e\), is p/q a terminating decimal?

(1) a > c

(2) b > d

IMO B..

Explanation:

\(\frac{p}{q} = \frac{2^a*3^b}{2^c*3^d*5^e}\)

For any fraction to be terminating... the denominator should be in form of \(2^m*5^n\) in it's lowest form where m and n are non negative integers - could be 0 also...

1. a > c Therefore a-c (let say k) > 0

Hence: \(\frac{p}{q} = \frac{2^k*3^b}{3^d*5^e}\)... which is could be a terminating decimal if b>d... or non terminating... if b<d. INSUFF...

For example the fraction could be .... \(\frac{4*3}{2*9*5}=0.13333...\) or \(\frac{4*9}{2*3*5}=0.12\)

2. b>d Therefore b-c (let say n) > 0 Hence: \(\frac{p}{q} = \frac{2^a*3^n}{2^c*5^e}\)... This is clearly a terminating decimal as the denominator would be in a form of \(2^m*5^n\)

For example the fraction could be .... \(\frac{4*3}{2*5}=0.12\) or \(\frac{4*3}{8*5}=0.3\)
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Re: Any decimal that has only a finite number of nonzero digits [#permalink]

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25 May 2012, 20:14

picked B. knew that 2/5 in the denominator will lead to a terminating decimal irrespective of a numerator. However didnt know of the formal rule. very helpful.

Re: Any decimal that has only a finite number of nonzero digits [#permalink]

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13 Jun 2012, 23:41

My answer is B. What's the OA?

The only determinant if p/q is a terminating decimal is 3^b/3^d since a fraction with the format a/b is terminal if b is a power of 2 or 5 or both.
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Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 36, 0.72, and 3.005 are terminating decimals.

If a, b, c, d and e are non-negative integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal?

(1) a > c

(2) b > d

Merging similar topics. Please refer to the solutions above.

Bunuel,

if denominator has only 2 or only 5, it seems that the fraction will also be a terminating decimal. Right?

Yes.

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.

Notice that when n or m equals to zero then the denominator will have only 2's or 5's.

The only determinant if p/q is a terminating decimal is 3^b/3^d since a fraction with the format a/b is terminal if b is a power of 2 or 5 or both.

OA is given in the initial post and it's B.

Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 36, 0.72, and 3.005 are terminating decimals. If a, b, c, d and e are non-negative integers and p=2^a*3^b and q=2^c*3^d*5^e, is p/q a terminating decimal?

So, according to that \(\frac{p}{q}=\frac{2^a*3^b}{2^c*3^d*5^e}\) will be terminating decimal if \(3^d\) in the denominator can be reduced, which will happen when the power of 3 in the numerator is more than or equal to the power of 3 in the denominator, so when \(b\geq{d}\).

As we can see (1) is completely irrelevant to answer whether \(b\geq{d}\), while (2) directly answers the question by stating that \(b>d\).

Re: Any decimal that has only a finite number of nonzero digits [#permalink]

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21 Jul 2012, 07:14

In the given problem, if a and c are not considered, then there may be a case when 2^a can be completely divided by 2^c, in that case, how the answer can be b? or if a-c is positive. Please let me know where am I thinking wrong?

In the given problem, if a and c are not considered, then there may be a case when 2^a can be completely divided by 2^c, in that case, how the answer can be b? or if a-c is positive. Please let me know where am I thinking wrong?

Not sure I understand what you mean above. What difference does it make whether 2^a is reduced or not? Or whether a-c is positive?

Re: Any decimal that has only a finite number of nonzero digits [#permalink]

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29 Jul 2012, 00:21

If the denominator in a fraction has only 2 or/and 5 --> terminating decimal If the denominator has any other prime factor other than 2 or 5 --> non-terminating decimal.

Hence is this question, we need to find out if 3 will be present in the denominator or not! . which means we need to find out if b > d . Hence OA : B.
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Re: Any decimal that has only a finite number of nonzero digits [#permalink]

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11 Aug 2012, 06:15

vscid wrote:

Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 36, 0.72, and 3.005 are terminating decimals. If a, b, c, d and e are non-negative integers and p=2^a*3^b and q=2^c*3^d*5^e, is p/q a terminating decimal?

A/B will be terminating(T) only if 1/B is T. B = 1/5^e will be always T , in the same way as 1/3^e will always be non- terminating(NT). Product of a NT with a T will always be NT.

In the numerator of p/q, powers of 2 and 3 can be +ve or -ve. Power of 2 doesn't have any affect on the T behaviour of p/q, but if power of 3 is -ve, it will go down to denominator and 1/3^x is always NT and make p/q NT.

stmt 1: a > c means power of 2 will be positive - INSUFFICIENT. stmt 2: b > d means power of 3 will be positive and 3^(b-d) will remain in the numerator. Thus, p/q will be T. SUFFICIENT

Answer is B
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