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# DS #241 Terminating Decimal - Another way to look at it?

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Manager
Joined: 30 Oct 2004
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DS #241 Terminating Decimal - Another way to look at it? [#permalink]  18 Dec 2004, 17:59
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Question Stats:

100% (01:36) correct 0% (00:00) wrong based on 1 sessions
Here's the original question:

Any decimal that has only a finite number of nonzero digits is a terminating decimal. ... If r and s are positive integers and the ratio r/s is expressed as a decimal, is r/s a terminating decimal?

1) 90<r<100

2) s=4

Answer and further question in the next post.
Manager
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Re: DS #241 Terminating Decimal - Another way to look at it? [#permalink]  18 Dec 2004, 18:04

Here's the partial explanation:

"Statement (2) alone is sufficient since division by the number 4 must terminate: the remainder when dividing by 4 must be 0, 1, 2, or 3, so the quotient must end with .0, .25, .5, or .75, respectively."

Is this true for r/s if s = any integer?

For instance, I tried the case when s = 5. The remainders are 0, 1, 2, 3, or 4. The quotient terminates in all cases.

Another way of looking at it is: Can anyone give an example of a similar question where the decimal doesn't terminate?
CIO
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Re: DS #241 Terminating Decimal - Another way to look at it? [#permalink]  18 Dec 2004, 18:09
Questor wrote:

Here's the partial explanation:

"Statement (2) alone is sufficient since division by the number 4 must terminate: the remainder when dividing by 4 must be 0, 1, 2, or 3, so the quotient must end with .0, .25, .5, or .75, respectively."

Is this true for r/s if s = any integer?

For instance, I tried the case when s = 5. The remainders are 0, 1, 2, 3, or 4. The quotient terminates in all cases.

Another way of looking at it is: Can anyone give an example of a similar question where the decimal doesn't terminate?

It would always be true with 5, as in the example above. Also 2, 8, 1, and 10. But it wouldn't work with 3, 6, or 9, and perhaps not 7 either. There are different patterns for different numbers, and it pays to be comfortable with all of them.

That said, I think it's a good question.
Director
Joined: 07 Nov 2004
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From the website
http://mathforum.org/library/drmath/view/58174.html

Quote:
The fraction will terminate if and only if the denominator has for
prime divisors only 2 and 5, that is, if and only if the denominator
has the form 2^a * 5^b for some exponents a >= 0 and b >= 0. The
number of decimal places until it terminates is the larger of a
and b.

The proof of this lies in the fact that every terminating decimal
has the form n/10^e, for some e >= 0 (e is the number of places to
the right that the decimal point must be moved to give you an integer,
and n is that integer), and every fraction of that form has a
terminating decimal found by writing down n and moving the decimal
point e places to the left. Now when you cancel common factors from
n/10^e = n/(2*5)^e = n/(2^e*5^e), it may reduce the exponents
in the denominator, but that is all that can happen.
Manager
Joined: 30 Oct 2004
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RE: [#permalink]  20 Dec 2004, 18:55
Great! Thanks!
RE:   [#permalink] 20 Dec 2004, 18:55
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