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Senior Manager  Joined: 19 Nov 2007
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Any decimal that has only a finite number of nonzero digits  [#permalink]

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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?

(1) a > c

(2) b > d

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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.

Hope it helps.
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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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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.
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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.

Correct me if I am missing some thing here ..!
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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
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1. Not sufficient
As 3 will be there in denominator or not.

2.sufficient
As we know 3 is not in the denominator and denominator has a 5.

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

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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.
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1
Hi,

$$p = 2^a3^b$$ and $$q = 2^c3^d5^e$$
$$p/q = 2^(a-c)3^(b-d)5^(-e)$$

division by 2 or 5 (raised to any power will result to terminating decimal)

only check is on b-d

so, using (2)
b-d > 0,
thus 3 is not in denominator, hence p/q is terminating decimal.

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

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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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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.

Hope it's clear.
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Re: Any decimal that has only a finite number of nonzero digits  [#permalink]

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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?

(1) a > c
(2) b > d

Check this: any-decimal-that-has-only-a-finite-number-of-nonzero-digits-90504.html#p689656

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 $$b>d$$.

Hope it's clear.
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Re: Any decimal that has only a finite number of nonzero digits  [#permalink]

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

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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?

Solution: any-decimal-that-has-only-a-finite-number-of-nonzero-digits-90504.html#p1096500
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Re: Any decimal that has only a finite number of nonzero digits  [#permalink]

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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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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?

(1) a > c

(2) b > d

p/q = (2^a * 3^b) / (2^c * 3^d * 5^e)
= [2^(a-c) * 3(b-d)] / 5^e

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

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

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Brunel -

Just to be 100% clear -- ANY fraction must have a denominator of 2, 5, or some multiple of the two numbers to terminate?
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Any decimal that has only a finite number of nonzero digits  [#permalink]

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lpetroski wrote:
Brunel -

Just to be 100% clear -- ANY fraction must have a denominator of 2, 5, or some multiple of the two numbers to terminate?

Yes. Fractions with denominators multiples of 2 or 5 or 2 AND 5 ONLY will give you a terminating decimal representation.

Examples. x/25, x/100, x/125, x/200 etc.
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Re: Any decimal that has only a finite number of nonzero digits  [#permalink]

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My approach to solve this problem:
p/q = (2^(a-c)*3^(b-d))/5^e
statement 1) a>c.
=> the 2^(a-c) is an integer, more so , since 2 is positive, it will return a positive integer.
case1) b>d then 3^(b-d) is an integer . Now whatever be the value of e (again integer) we will have a denominator with powers of 5. We know that if we divide by 5 or its powers, we always get a terminating decimal. The reason is that we are always multiplying 5 with itself. To see this logic. let us say we are dividing a number by 5^2. Simply divide the number by the first 5. This always gives a terminating decimal. Then divide that terminating decimal number with another 5 remaining in the denominator. This logic can be extended to any powers of 5.

case 2) let us try to reduce this to 2/3 ..it is possible to set b-d = 1 and e=0 and a-c=1. We know 2/3 is non terminating
Therefore insuff

statement #2
b>d:- This guarentees that 3^(b-d) is an integer.
Now if a> or =c and e>0 this implies that we will divide an integer by 5 .. answer is terminating decimal
now if a <c and e >0 this implies that we divide integer by 10, answer is yes
what if e< 0 then either we will have p/q as integer or a number divisible by 2 .. we know this will also be a terminating decminal

So statement #2 is sufficient -
B Re: Any decimal that has only a finite number of nonzero digits   [#permalink] 12 May 2016, 19:23

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