What is a terminating decimal?
There is a question in MGMAT 6 about them. It sais:
For fraction p/q to be a terminating decimal, the numerator must be an integer and the denominator must be an integer that can be expressed in the form of 2^x 5^y where x and y are nonnegative integers. (Any integer divided by a power of 2 or 5 will result in a terminating decimal.)
Can anyone explain with an example because I dont see it clear.
Thanks
GMAT Forum
- Luci
Terminating decimal. MGMAT EX6.Quest 16
Sorry this is the problem I forgot to type it:
If a, b, c, d and e are integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal?
(1) a > c
(2) b > d
If a, b, c, d and e are integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal?
(1) a > c
(2) b > d
- harish.dorai
Luci:
A terminating decimal is something which will have a finite number of digits after the decimal point.
Ex: 1.5 is a terminating decimal which is also the fraction 3/2. Where as if we express the fraction 4/3, it will be a non-terminating decimal which equals 1.3333333...... Another type of non-terminating decimal is where a group of digits repeats. Ex: 1.232323.....
There are some interesting non-terminating such as the constant PI which you encounter in formula for the area and circumference of circles. Its value is 3.1415982...etc, (Here the digits are random but they don't terminate).
For a decimal to terminate, the underlying fraction that defines the decimal should meet any of the following conditions.
1) The fraction should have a numerator of 2, 4, 8, 16 etc. OR all non-negative powers of 2. (Ex: 2^2 = 4, 2^3 = 8)
Ex: 1/2 = 0.5, 5/4 = 1.25, 18/8 = 2.25
2) The fraction has to be divisible by 5 or 5^x (all non-negative powers of 5)
Ex: 17/5 = 3.4, 180/25 = 7.2 (remember 25 is 5^2)
3) The fraction should be divisible by 10 or 10^x (all non-negative powers of 10)
Ex: 1893/1000 = 1.893, 176/100 = 1.76
Now if you look at statement (3), 10 is actually composed of 2 and 5. So 10^x can be expressed 2^x multiplied by 5^x.
And if you look at 2^2. You can say that as 2^2 multiplied by 5^0(which is 1). Similarly 5^3 can be said as 5^3 multiplied by 2^0 (which is 1). So if you combine the above 3 conditions we can come up with a general conclusion that for a number to be a terminating decimal, its fraction equivalent should have the denominator in the form 2^x multiplied by 5^y.
Now coming to your problem, it says that p = 2^a3^b and q = 2^c3^d5^e. So in order for p/q to be a terminating decimal, the trouble maker here is the 3^d that is present in q. We need to find out if we can get rid of that. If we can get rid of 3^d, we will get the denominator in the 2^x multiplied by 5^y format. Now let us examine the statements. I will start off with statement (2).
Statement (2) - It says that b > d. So if you write p/q, since 3^b is in the numerator and 3^d is in the denominator and b > d, p/q will have 3^(something in the denominator). So it violates the requirement for a terminating decimal and we can say that p/q is not a terminating decimal. SUFFICIENT.
Statement (1) - It says a > c, which are the exponents of 2. It doesn't help us anyway to deal with the exponents of 3. So this is INSUFFICIENT.
The answer is (B).
A terminating decimal is something which will have a finite number of digits after the decimal point.
Ex: 1.5 is a terminating decimal which is also the fraction 3/2. Where as if we express the fraction 4/3, it will be a non-terminating decimal which equals 1.3333333...... Another type of non-terminating decimal is where a group of digits repeats. Ex: 1.232323.....
There are some interesting non-terminating such as the constant PI which you encounter in formula for the area and circumference of circles. Its value is 3.1415982...etc, (Here the digits are random but they don't terminate).
For a decimal to terminate, the underlying fraction that defines the decimal should meet any of the following conditions.
1) The fraction should have a numerator of 2, 4, 8, 16 etc. OR all non-negative powers of 2. (Ex: 2^2 = 4, 2^3 = 8)
Ex: 1/2 = 0.5, 5/4 = 1.25, 18/8 = 2.25
2) The fraction has to be divisible by 5 or 5^x (all non-negative powers of 5)
Ex: 17/5 = 3.4, 180/25 = 7.2 (remember 25 is 5^2)
3) The fraction should be divisible by 10 or 10^x (all non-negative powers of 10)
Ex: 1893/1000 = 1.893, 176/100 = 1.76
Now if you look at statement (3), 10 is actually composed of 2 and 5. So 10^x can be expressed 2^x multiplied by 5^x.
And if you look at 2^2. You can say that as 2^2 multiplied by 5^0(which is 1). Similarly 5^3 can be said as 5^3 multiplied by 2^0 (which is 1). So if you combine the above 3 conditions we can come up with a general conclusion that for a number to be a terminating decimal, its fraction equivalent should have the denominator in the form 2^x multiplied by 5^y.
Now coming to your problem, it says that p = 2^a3^b and q = 2^c3^d5^e. So in order for p/q to be a terminating decimal, the trouble maker here is the 3^d that is present in q. We need to find out if we can get rid of that. If we can get rid of 3^d, we will get the denominator in the 2^x multiplied by 5^y format. Now let us examine the statements. I will start off with statement (2).
Statement (2) - It says that b > d. So if you write p/q, since 3^b is in the numerator and 3^d is in the denominator and b > d, p/q will have 3^(something in the denominator). So it violates the requirement for a terminating decimal and we can say that p/q is not a terminating decimal. SUFFICIENT.
Statement (1) - It says a > c, which are the exponents of 2. It doesn't help us anyway to deal with the exponents of 3. So this is INSUFFICIENT.
The answer is (B).
- StaceyKoprince
- ManhattanGMAT Staff
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- Luci
Thanks
Thanks Haris, nice explanation. Hopefully I wont fail a problem about terminating decimals now
- Guest
Doubt
Harish
I have a doubt. in the second last para you have mentioned:
Statement (2) - It says that b > d. So if you write p/q, since 3^b is in the numerator and 3^d is in the denominator and b > d, p/q will have 3^(something in the denominator). So it violates the requirement for a terminating decimal and we can say that p/q is not a terminating decimal. SUFFICIENT.
However, say b= 5 and d=3; since b>d
then 3^5/ 3^3, will result in 3^2 in numerator, not denominator!!
You have mentioned that p/q will have 3^(something in the denominator)??
The answer would remain same, (A), but it doesn't violate the requirement for a terminating decimal and we can say that p/q is a terminating decimal.
Plz confirm your views about the query above.
I have a doubt. in the second last para you have mentioned:
Statement (2) - It says that b > d. So if you write p/q, since 3^b is in the numerator and 3^d is in the denominator and b > d, p/q will have 3^(something in the denominator). So it violates the requirement for a terminating decimal and we can say that p/q is not a terminating decimal. SUFFICIENT.
However, say b= 5 and d=3; since b>d
then 3^5/ 3^3, will result in 3^2 in numerator, not denominator!!
You have mentioned that p/q will have 3^(something in the denominator)??
The answer would remain same, (A), but it doesn't violate the requirement for a terminating decimal and we can say that p/q is a terminating decimal.
Plz confirm your views about the query above.
- StaceyKoprince
- ManhattanGMAT Staff
- Posts: 9360
- Joined: Wed Oct 19, 2005 9:05 am
- Location: Montreal
Yep - probably just reversed his thinking by accident in that very long explanation! But, yes, if b>d, then the 3 will drop out of the denominator and we'll just have 3^something in the numerator. As the guest noted, since it's a yes/no question, this doesn't actually change the outcome.
Stacey Koprince
Instructor
Director, Content & Curriculum
ManhattanPrep
Instructor
Director, Content & Curriculum
ManhattanPrep
- unique
Anonymous Wrote:Sorry for the mistake. Though it doesn't change the answer. It should leave 3^(Something) in the Numerator. So that will leave 2^c and 5^d in the denominator which satisfies the condition for a terminating decimal.
Thats true. But where does the non-negative part come into picture. Because in statement 2 we have 2^c x 5^e in the denominator.
Both c and e can be negative. If a>c then p/q won't be a non-terminating decimal but an it will result in an integer.
example -
b>d b = 3 d = 2 a = -1 c = -4 e = -1
This will result in 2 ^3 x 3 x 5
- unique
unique Wrote:Anonymous Wrote:Sorry for the mistake. Though it doesn't change the answer. It should leave 3^(Something) in the Numerator. So that will leave 2^c and 5^d in the denominator which satisfies the condition for a terminating decimal.
Thats true. But where does the non-negative part come into picture. Because in statement 2 we have 2^c x 5^e in the denominator.
Both c and e can be negative. If a>c then p/q won't be a non-terminating decimal but an it will result in an integer.
example -
b>d b = 3 d = 2 a = -1 c = -4 e = -1
This will result in 2 ^3 x 3 x 5
Just to be clear - I got answer E.
non-negative power of 2 or 5 is not guaranteed in denominator by either statement or together. Did I miss something?
- Harish Dorai
If you have negative powers in the denominator, then wouldn't that be equivalent to positive powers in the numerator?
Example: 1/(2 to the power -3) is actually 2 to the power 3 which is 8.
So I think we are fine, even with negative powers for 2 and 5 in the denominator. My previous explanation might need some correction.
Example: 1/(2 to the power -3) is actually 2 to the power 3 which is 8.
So I think we are fine, even with negative powers for 2 and 5 in the denominator. My previous explanation might need some correction.
- christiancryan
- Course Students
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Yes, to answer your question, unique -- those are both examples of terminating decimals.
The question raised earlier about whether the powers of 2 and 5 in the denominator can be non-negative can be resolved this way, I think. First of all, you have to assume that you've already reduced the numerator and the denominator to lowest terms (after all, you could have a lurking 3 in both the numerator and denominator, e.g. 3/30, but that IS a terminating decimal because *after you've canceled,* you're left with an integer on top and an integer on the bottom that is expressible as a product of non-negative, integer powers of 2 and 5). Along the same lines, you should assume that before you apply the "non-negative integer powers of 2 and 5" test to the denominator, you already MOVE every term where it belongs. In other words, if you have a negative power of 2 in the denominator right now, move it to the numerator and make it a positive power.
The test is this: reduce the fraction to lowest terms and express both the numerator and the denominator as integers. If you're not sure where a particular term goes, because its power could be positive or negative, then you can set up cases. For instance, 3^x / 2^x is ambiguous, because we don't know the sign of x (or whether it's an integer, for that matter.
To be terminating, *after reducing* the numerator must be an integer (any integer), and the denominator must be an integer expressible as some nonnegative, integer power of 2, times some nonnegative, integer power of 5. In other words, the denominator's prime factorization must ONLY contain 2's and 5's. The reason is that only under these conditions can you reexpress the fraction as another integer over a power of 10, by multiplying top & bottom by an appropriate integer.
E.g., 1/4 = (1*25)/(4*25) = 25/100 = 0.25
A terminating decimal can in fact be defined as a decimal expressible as an integer over a power of 10 (a nonnegative integer power, that is!).
Hope this helps!
The question raised earlier about whether the powers of 2 and 5 in the denominator can be non-negative can be resolved this way, I think. First of all, you have to assume that you've already reduced the numerator and the denominator to lowest terms (after all, you could have a lurking 3 in both the numerator and denominator, e.g. 3/30, but that IS a terminating decimal because *after you've canceled,* you're left with an integer on top and an integer on the bottom that is expressible as a product of non-negative, integer powers of 2 and 5). Along the same lines, you should assume that before you apply the "non-negative integer powers of 2 and 5" test to the denominator, you already MOVE every term where it belongs. In other words, if you have a negative power of 2 in the denominator right now, move it to the numerator and make it a positive power.
The test is this: reduce the fraction to lowest terms and express both the numerator and the denominator as integers. If you're not sure where a particular term goes, because its power could be positive or negative, then you can set up cases. For instance, 3^x / 2^x is ambiguous, because we don't know the sign of x (or whether it's an integer, for that matter.
To be terminating, *after reducing* the numerator must be an integer (any integer), and the denominator must be an integer expressible as some nonnegative, integer power of 2, times some nonnegative, integer power of 5. In other words, the denominator's prime factorization must ONLY contain 2's and 5's. The reason is that only under these conditions can you reexpress the fraction as another integer over a power of 10, by multiplying top & bottom by an appropriate integer.
E.g., 1/4 = (1*25)/(4*25) = 25/100 = 0.25
A terminating decimal can in fact be defined as a decimal expressible as an integer over a power of 10 (a nonnegative integer power, that is!).
Hope this helps!
- UPA
DWG Wrote:I am still confused though.
What about a fraction like 43/256?
This fraction, I believe, is in lowest terms and the denominator can be expressed as a power of 2. Thus that would indicate that it terminates, when in actuality it does not.
Can someone explain this to me?
Thanks.
43/256 = 43 (.5)^8
it is a terminating decimal.