Here is a question:
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
Can you please provide an answer to this?
Terminating Decimals
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- Ian Stewart
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You can recognize whether a fraction will produce a terminating decimal by:chrisjim5 wrote:Here is a question:
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
Can you please provide an answer to this?
1. Reducing your fraction completely
2. Then looking only at the prime factors of the denominator. If the denominator has any prime factor besides 2 or 5, the fraction will give a *repeating* (infinite) decimal. If the only prime factors of the denominator are 2 and/or 5, the fraction will give a terminating decimal.
So fractions like 3/16 (only prime factor of denominator is 2), 9/125 (only prime factor of denominator is 5) and 3/40 (only prime factors of denominator are 2 and 5) will all produce terminating decimals. Fractions like 1/13, 9/35, and 11/120 will all produce non-terminating decimals since each is completely reduced, and has a factor different from 2 or 5 in the denominator. The first step above is critical; while a fraction like 7/35 might appear to have a factor of 7 in the denominator, that 7 actually cancels with the 7 in the numerator to give us 1/5, a terminating decimal.
So in this question, we have the fraction:
(2^a*3^b) / (2^c*3^d*5^e)
The only reason this might not terminate is because of the 3's; if our 3^d in the denominator does not cancel out completely, we will get a repeating decimal. If it does cancel, we will get a terminating decimal. Statement 2 tells us that it will cancel completely, so is sufficient. Statement 1 doesn't help. So the answer is B.
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- fskilnik@GMATH
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The solution provided by Ian is perfect and very informative, but I guess many test takers go to the exam without this knowledge... that´s why I believe the solution I present below is also nice:
From the "a, b, c, d and e are non-negative integers" hypothesis, I believe the question should be "seen" as:
2^(a-c) * 3^(b-d) over 5^e is a terminating decimal ?
Important: this is just a visually helpful thing, because (for instance) the value of (a-c) may be negative, and that means that 2^(a-c) can be at the denominator, "in reality"...
(1) This sttm tells us that 2 is really in numerator, but what about the 3(´s) ?
> Take a = 2, c =1 (to be only 2^1) and b = 2 and d = 1 (to let 3 be 3^1, so numerator...) and e = 1 , then we have: (2 * 3) over 5, and if you multiply both numerator and denumerator by 2 , you get (2^2 * 3) over... 10, that is, certainly terminating because it is an integer divided by 10, so you just move decimal point, you do not "alter terminallity"...
> Take a =2 , c = 1 (to be only 2^1 again) and b = 1 and d = 2 (to let 3 be 3^(-1), so denumerator) and e =1 , then we have: 2 over (3 times 5) and now we know we are with a non-terminating decimal because of this (for instance):
2/(15) = 2*2 / (15*2) = 4/30 = (1/10) * (4/3) and divide by 10 does not alter the "terminallity of a certain decimal" (as mentioned above) and we know that 4/3 is not terminating, because it is equal to 1+ 1/3 and (1 is an integer and) 1/3 is non-terminating, for sure (0.333333...)
Obs.: this is not stupid calculations, I believe. This is the "insight" that is behind Ian´s statements...
(2) Now we know that 3´s are on the numerator, therefore we may have only 2´s , only 5´s or both in the denumerator. From all shown, you should be able to recognize that 2´s and 5´s are no problem, because multiplying by 10´s in enough quantity you turn the fractions into integers, therefore terminating decimals.
This one DECIDES affirmatively on the question asked, that is, (2) is sufficient.
Regards,
Fabio.
From the "a, b, c, d and e are non-negative integers" hypothesis, I believe the question should be "seen" as:
2^(a-c) * 3^(b-d) over 5^e is a terminating decimal ?
Important: this is just a visually helpful thing, because (for instance) the value of (a-c) may be negative, and that means that 2^(a-c) can be at the denominator, "in reality"...
(1) This sttm tells us that 2 is really in numerator, but what about the 3(´s) ?
> Take a = 2, c =1 (to be only 2^1) and b = 2 and d = 1 (to let 3 be 3^1, so numerator...) and e = 1 , then we have: (2 * 3) over 5, and if you multiply both numerator and denumerator by 2 , you get (2^2 * 3) over... 10, that is, certainly terminating because it is an integer divided by 10, so you just move decimal point, you do not "alter terminallity"...
> Take a =2 , c = 1 (to be only 2^1 again) and b = 1 and d = 2 (to let 3 be 3^(-1), so denumerator) and e =1 , then we have: 2 over (3 times 5) and now we know we are with a non-terminating decimal because of this (for instance):
2/(15) = 2*2 / (15*2) = 4/30 = (1/10) * (4/3) and divide by 10 does not alter the "terminallity of a certain decimal" (as mentioned above) and we know that 4/3 is not terminating, because it is equal to 1+ 1/3 and (1 is an integer and) 1/3 is non-terminating, for sure (0.333333...)
Obs.: this is not stupid calculations, I believe. This is the "insight" that is behind Ian´s statements...
(2) Now we know that 3´s are on the numerator, therefore we may have only 2´s , only 5´s or both in the denumerator. From all shown, you should be able to recognize that 2´s and 5´s are no problem, because multiplying by 10´s in enough quantity you turn the fractions into integers, therefore terminating decimals.
This one DECIDES affirmatively on the question asked, that is, (2) is sufficient.
Regards,
Fabio.
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The ans is B
when we simplify the question we get 2^(a-c) * 3^(b-d) / 5^e
when we have prime factors other than 2 and 5 in the denominator we get non-terminating decimal.
1) a>c is in sufficient. If a>c or a<c it doesnt matter since we don't know about b and d.
2) if b>d there is no 3 in the denominator. So, we have terminating number.
Hence 2 alone is sufficient to ans the question.
when we simplify the question we get 2^(a-c) * 3^(b-d) / 5^e
when we have prime factors other than 2 and 5 in the denominator we get non-terminating decimal.
1) a>c is in sufficient. If a>c or a<c it doesnt matter since we don't know about b and d.
2) if b>d there is no 3 in the denominator. So, we have terminating number.
Hence 2 alone is sufficient to ans the question.
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mgmat cat question hmmmmmmmmm
already enough on this by the experts, my say in short.
remember 2,4,6,8 and 10(because it has 2 and 5) are terminators(the bad guys)(schwarzenegger in terminator 1)
any fraction with additional 3,6(because it has 3),7 and 9(because it has 3) as denominators is not a terminator and the good guy(schwarzenegger in terminator 2)
hope this helps, u'll remember this now
this question can simply be rephrased as
is the no of 3's in p greater than no of 3's in q ???
i'e
is b>d
the answer is right in one of the statements 8)
already enough on this by the experts, my say in short.
remember 2,4,6,8 and 10(because it has 2 and 5) are terminators(the bad guys)(schwarzenegger in terminator 1)
any fraction with additional 3,6(because it has 3),7 and 9(because it has 3) as denominators is not a terminator and the good guy(schwarzenegger in terminator 2)
hope this helps, u'll remember this now
this question can simply be rephrased as
is the no of 3's in p greater than no of 3's in q ???
i'e
is b>d
the answer is right in one of the statements 8)
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p/q = 2^(a-c)*3^(b-d)/5^e
Except b<d where 3 comes in the denominator in all other cases the fraction is a terminating decimal
Statement 1: INSUFFICIENT
Statement 2: SUFFICIENT
Except b<d where 3 comes in the denominator in all other cases the fraction is a terminating decimal
Statement 1: INSUFFICIENT
Statement 2: SUFFICIENT
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if the fraction has powers of 2 or 5 only hen the fraction is terminating decimal
S1: a>c does not tell about power of 3
S2: b-d>0 hence the fraction is terminating decimal
(B) is ans
S1: a>c does not tell about power of 3
S2: b-d>0 hence the fraction is terminating decimal
(B) is ans
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Nice Logic Ian.Thanks!!Ian Stewart wrote:You can recognize whether a fraction will produce a terminating decimal by:chrisjim5 wrote:Here is a question:
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
Can you please provide an answer to this?
1. Reducing your fraction completely
2. Then looking only at the prime factors of the denominator. If the denominator has any prime factor besides 2 or 5, the fraction will give a *repeating* (infinite) decimal. If the only prime factors of the denominator are 2 and/or 5, the fraction will give a terminating decimal.
So fractions like 3/16 (only prime factor of denominator is 2), 9/125 (only prime factor of denominator is 5) and 3/40 (only prime factors of denominator are 2 and 5) will all produce terminating decimals. Fractions like 1/13, 9/35, and 11/120 will all produce non-terminating decimals since each is completely reduced, and has a factor different from 2 or 5 in the denominator. The first step above is critical; while a fraction like 7/35 might appear to have a factor of 7 in the denominator, that 7 actually cancels with the 7 in the numerator to give us 1/5, a terminating decimal.
So in this question, we have the fraction:
(2^a*3^b) / (2^c*3^d*5^e)
The only reason this might not terminate is because of the 3's; if our 3^d in the denominator does not cancel out completely, we will get a repeating decimal. If it does cancel, we will get a terminating decimal. Statement 2 tells us that it will cancel completely, so is sufficient. Statement 1 doesn't help. So the answer is B.
:arrow: If the denominator has any prime factor besides 2 or 5, the fraction will give a *repeating* (infinite) decimal. If the only prime factors of the denominator are 2 and/or 5, the fraction will give a terminating decimal.
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Relatively easier approach for you.Rastis wrote:Can someone provide and easier to understand explanation please?
Lets put this question as 2^(a-c).3^(b-d)/5^e.
Its for sure that this question wishes us to predict nature of 2^(a-c).3^(b-d)/5^e by deducing nature of 2^(a-c), 3^(b-d), & 5^e.
=> lets put a=c=b=d=0. This will help us find the nature of 5^e.
=> so 2^(a-c).3^(b-d)/5^e = 2^0.3^0/5^e = 1/5^e.
=> Now you may try few values of e such as 1, 2, 3... to see if 1/5^e is terminating or non-terminating. We can conclude 1/5^e is terminating.So 5^e has no role to decide if expression is non-terminating.
Now lets put b=d=e=0. This will help us find the nature of 2^(a-c).
=> so 2^(a-c).3^(b-d)/5^e = 2^(a-c).3^0/5^0 = 2^(a-c).
=> Now you may try few values of (a-c) such as ...., -3, -2, -1 to see if 2^(a-c) is terminating or non-terminating. We can conclude 2^(a-c) is terminating.So 2^(a-c) has no role to decide if expression is non-terminating. Whether a > < c. So stat. 1 is not necessary. Redundant.
Now lets put a=c=e=0. This will help us find the nature of 3^(b-d).
=> so 2^(a-c).3^(b-d)/5^e = 2^0.3^(b-d)/5^0 = 3^(b-d).
=> Now you may try few values of(b-d) such as ...., -3, -2, -1 to see if 3^(b-d) is terminating or non-terminating. We can conclude 3^(b-d) is non-terminating if (b-d) is negative.So to make it terminating (b-d) should be positive or b>d. So stat. 2 is not necessary & sufficient.
Ans B.
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My explanation is similar to the ones above, but perhaps you will find it helpful:Rastis wrote:Can someone provide and easier to understand explanation please?
https://www.beatthegmat.com/terminating- ... 92476.html
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