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Re: If P and Q are positive integers, and n is the decimal [#permalink]
20 Jul 2013, 09:51

A. P/Q = (49/256)= (7)(7)/(16)(16) P/Q = (7/16)*(7/16) = .4125 * .4125 = finite B. 32 = 2^5. Any number (odd/even) divided by 2^n will always be finite. C. 75/384 = (3*5^2)/(2^7*3) ---> 3 gets cancelled and we have 5^2 / 2^7 - always finite because of 2^7.

Re: If P and Q are positive integers, and n is the decimal [#permalink]
20 Jul 2013, 22:49

Expert's post

The best trick to find out whether a fraction will yield a definite decimal number is to check whether the denominator can be expressed in terms of the power of 2 and/or 5. If yes, then the fraction will yield a definite decimal. In the above question 256, 32 and 384 can be expressed in powers of 2 as well. Hence I, II and III are correct. Regards _________________

Re: If P and Q are positive integers, [#permalink]
21 Jul 2013, 00:42

Expert's post

avinashrao9 wrote:

MacFauz wrote:

Jp27 wrote:

My doubt is if it were given P and Q to be positive numbers and I)& III) are only correct right? As the P can be 1/3.

Cheers

] I should think so... Infact.. If it had been given as postive numbers, P could be any irrational number such as \sqrt{2},\sqrt{3}, \sqrt{5}

So, the answer would be only 1 & 3.

Kudos Please... If my post helped.

As long as the denominator can be expressed as powers of prime factors, the fraction will always be finite...

That's not true. Any positive integer can be expressed as powers of primes.

Theory: Reduced fraction \frac{a}{b} (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and onlyb (denominator) is of the form 2^n5^m, where m and n are non-negative integers. For example: \frac{7}{250} is a terminating decimal 0.028, as 250 (denominator) equals to 2*5^2. Fraction \frac{3}{30} is also a terminating decimal, as \frac{3}{30}=\frac{1}{10} and denominator 10=2*5.

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \frac{x}{2^n5^m}, (where x, n and m are integers) will always be the terminating decimal.

We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \frac{6}{15} has 3 as prime in denominator and we need to know if it can be reduced.

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