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radhika1306
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 Posted: Tue Sep 04, 2007 12:03 pm Post subject: set 24 Q 26 |
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Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only
a finite number of nonzero digits?
(1) P>Q
(2) Q=8 |
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ri2007
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| Posted: Tue Sep 04, 2007 12:51 pm Post subject: Re: set 24 Q 26 |
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| radhika1306 wrote: | Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only
a finite number of nonzero digits?
(1) P>Q
(2) Q=8 |
I feel the answer should be B
If you consider statement 1 alone P can be 23 and Q can be 3. This will produe a recurring decimal number. However if P is 21 and Q is 3 it will contain only finite numbers of non zero digits.
Now consider statement two - The denominator is 8. To produce a recurring number the denominator should be a prime number ex - 2 and 5. See rule below. So for any value of P, P/Q will be a finite number.
I found this rule on the net:
A fraction in lowest terms with a prime denominator other than 2 or 5 (i.e. coprime to 10) always produces a recurring decimal. The period of the recurring decimal, 1/ p, where p is prime, is either p − 1 (the first group) or a divisor of p − 1 (the second group)
Please confirm if my logic is correct. |
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radhika1306
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| Posted: Tue Sep 04, 2007 2:09 pm Post subject: |
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| OA is E,can anyone help us |
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ri2007
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| Posted: Tue Sep 04, 2007 2:38 pm Post subject: |
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| radhika1306 wrote: | | OA is E,can anyone help us |
I dont understand this, after seeing the answer posted by you I even calculated the value of P/Q with Q =8 and P = all numbers from 1 to 7 and we always get a finite number. For any value over 8 the same pattern should repeat as the remander can only be any number from 0 to 8.
May be I misunderstood the question?? |
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radhika1306
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| Posted: Thu Sep 06, 2007 6:49 am Post subject: |
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| i thought the same, maybe the answer in the ans set is wrong |
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tanyajoseph
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| Posted: Mon Sep 10, 2007 2:31 pm Post subject: |
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NON ZERO DIGITS is the trap.
Unless you know P and Q you would not be able to really say that.So ans should be E. |
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TT
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| Posted: Mon Sep 10, 2007 3:03 pm Post subject: TT |
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| How is non-zero digit a trap? |
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samirpandeyit62
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| Posted: Tue Sep 11, 2007 1:29 am Post subject: |
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If the ans is E I think it must be incorrect or the Q has a typo
IMO ans should be B
coz it is mentioned that P & Q are both +ve integers
we have P/8
i.e let P = x + a ( where x is a multiple of 8 i.e qoutient when p is divided by 8 and a is the remainder)
so p/8 = (x+a)/8 i.e. x/8 + a/8
now x is the quoitient i.e divisible by 8 so the decimal is generated by a/8
i.e remainder/8
now a can be 1-7
try dividing 1-7 by 8 you will always get a decimal with finite nos after decimal pt.
Also as RI mentioned, 2 & 5 will divide any nos giving finite nos after decimal pt.
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Samir |
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| Posted: Tue Sep 11, 2007 8:10 am Post subject: Re: set 24 Q 26 |
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| radhika1306 wrote: | Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only
a finite number of nonzero digits?
(1) P>Q
(2) Q=8 |
It's B.
If you can represent denominator in the form of 2^m*5^n then fraction terminates.Furthermore, x/8 does not contain any zeros after decimal point. |
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