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Any decimal that has only a finite number of nonzero digits
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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 nonnegative 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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Re: terminating decimal!
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18 Feb 2010, 08:37
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 nonnegative 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 nonnegative 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 2s and/or 5s 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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Re: terminating decimal!
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20 Feb 2010, 03:10
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 nonnegative 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 ac (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 bc (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: terminating decimal!
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17 Feb 2010, 17:30
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 nonterminating definitively. Stmt1 deals with the relationship between a and c, so we know clearly that 2 will not be there in the denominator. Stmt2 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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Re: terminating decimal!
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17 Feb 2010, 17:46
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 nonterminating definitively.
Stmt1 deals with the relationship between a and c, so we know clearly that 2 will not be there in the denominator.
Stmt2 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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Re: terminating decimal!
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18 Feb 2010, 07:12
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 nonnegative 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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Re: terminating decimal!
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25 Apr 2011, 08:13
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.
Answer is B.
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Re: Any decimal that has only a finite number of nonzero digits
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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.



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Re: terminal decimal
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09 Jun 2012, 22:04
Hi,
\(p = 2^a3^b\) and \(q = 2^c3^d5^e\) \(p/q = 2^(ac)3^(bd)5^(e)\)
division by 2 or 5 (raised to any power will result to terminating decimal)
only check is on bd
so, using (2) bd > 0, thus 3 is not in denominator, hence p/q is terminating decimal.
Answer is (B).
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Re: terminal decimal
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13 Jun 2012, 23:37
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 nonnegative 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
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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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Re: terminal decimal
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14 Jun 2012, 00:31
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 nonnegative 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 nonnegative 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
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14 Jun 2012, 00:40
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 nonnegative 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: anydecimalthathasonlyafinitenumberofnonzerodigits90504.html#p689656So, 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 \(b>d\). Answer: B. Hope it's clear.
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Re: Any decimal that has only a finite number of nonzero digits
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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 ac 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
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22 Jul 2012, 03:32
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 ac 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 ac is positive? Please read this: anydecimalthathasonlyafinitenumberofnonzerodigits90504.html#p689656 and this: anydecimalthathasonlyafinitenumberofnonzerodigits90504.html#p1096496 for theory. Solution: anydecimalthathasonlyafinitenumberofnonzerodigits90504.html#p1096500
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Re: Any decimal that has only a finite number of nonzero digits
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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 > nonterminating 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
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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 nonnegative 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^(ac) * 3(bd)] / 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^(bd) 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
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13 Jan 2016, 16:13
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
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13 Jan 2016, 19:09
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
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12 May 2016, 19:23
My approach to solve this problem: p/q = (2^(ac)*3^(bd))/5^e statement 1) a>c. => the 2^(ac) is an integer, more so , since 2 is positive, it will return a positive integer. case1) b>d then 3^(bd) 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 bd = 1 and e=0 and ac=1. We know 2/3 is non terminating Therefore insuff
statement #2 b>d: This guarentees that 3^(bd) 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
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