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Any decimal that has only a finite number of nonzero digits [#permalink]
17 Feb 2010, 15:58

6

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

62% (02:02) correct
38% (01:25) wrong based on 240 sessions

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?

Re: terminating decimal! [#permalink]
17 Feb 2010, 16:46

BarneyStinson wrote:

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.

Re: terminating decimal! [#permalink]
18 Feb 2010, 06:12

vscid wrote:

nitishmahajan wrote:

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 = \(2^a3^b\) 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

Re: terminating decimal! [#permalink]
18 Feb 2010, 07:37

5

This post received KUDOS

Expert's post

vscid wrote:

nitishmahajan wrote:

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 = \(2^a3^b\) 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.

Re: terminating decimal! [#permalink]
20 Feb 2010, 02:10

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 = \(2^a3^b\) 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]
25 May 2012, 19:14

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]
13 Jun 2012, 22:41

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. _________________

Far better is it to dare mighty things, to win glorious triumphs, even though checkered by failure... than to rank with those poor spirits who neither enjoy nor suffer much, because they live in a gray twilight that knows not victory nor defeat. - T. Roosevelt

Re: terminal decimal [#permalink]
13 Jun 2012, 23:31

Expert's post

Rice wrote:

Bunuel wrote:

alchemist009 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^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.

Re: Any decimal that has only a finite number of nonzero digits [#permalink]
13 Jun 2012, 23:40

Expert's post

gmatsaga wrote:

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.

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]
21 Jul 2012, 06:14

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?

Re: Any decimal that has only a finite number of nonzero digits [#permalink]
22 Jul 2012, 02:32

Expert's post

pavanpuneet wrote:

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]
28 Jul 2012, 23:21

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]
11 Aug 2012, 05:15

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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Re: Any decimal that has only a finite number of nonzero digits [#permalink]
25 Jun 2014, 05:05

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