set 24 Q 26

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Re: set 24 Q 26

by ri2007 » Tue Sep 04, 2007 12:51 pm
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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by radhika1306 » Tue Sep 04, 2007 2:09 pm
OA is E,can anyone help us

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by ri2007 » Tue Sep 04, 2007 2:38 pm
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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by radhika1306 » Thu Sep 06, 2007 6:49 am
i thought the same, maybe the answer in the ans set is wrong

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by tanyajoseph » Mon Sep 10, 2007 2:31 pm
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

by TT » Mon Sep 10, 2007 3:03 pm
How is non-zero digit a trap?

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by samirpandeyit62 » Tue Sep 11, 2007 1:29 am
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.
Regards
Samir