If p and q are positive integers, is 21^p/630^q a terminating

Question

If p and q are positive integers, is  \frac{(21)^p}{(630)^q}  a terminating decimal

(1) p < 2q
(2) p > q

Source: Experts Global

DavidTutorexamPAL · Answer

We'll simplify the expression so we understand what we need to do.
This is a Precise approach.

We'll extract all common factors between 21 and 630, creating an easier-to-understand expression.
 \frac{(21)^p}{(630)^q}=\frac{(21)^p}{(21*30)^q}=\frac{(21)^{p-q}}{(30)^q}=\frac{(3*7)^{p-q}}{(3*10)^q}=\frac{(3)^{p-2q}*(7)^{p-q}}{(10)^q} 
A terminating decimal is an integer divided by some power of 10.
Therefore, for the above to be a terminating decimal,  (3)^{p-2q}*(7)^{p-q}  must be an integer.
As 3 and 7 are both prime numbers then this occurs only when p-2q and p-q are both positive. This happens only when p>2q.
Looking at our statements (1) lets us answer the question with a NO and (2) does not let us answer the question.

Our answer is (A).

*Note 1 - a terminating decimal is basically one long integer with the decimal point somewhere in the middle.
Therefore you can multiply it by a power of 10 to create an integer. This also means that every terminating decimal is an integer divided by a power of 10.
*Note 2 - A number divided by another number is an integer only if the denominator has the same prime factors as the numerator.
As this can never happen for powers of 3 and 7, they must both be in the numerator meaning that p-2q and p-q need to be positive.

TestPrepUnlimited · Answer

To have a terminating decimal, we need the denominator to only have factors of 2 or 5.

630^q = 63^q * 10^q = 7^q * 3^{2q} * 10^q . We can ignore  10^q , the 7 and 3's are the troublemakers and we need the numerator to cancel these out. The numerator  21^p = 7^p *3^p  has the responsibility of eliminating all the 3's in the denominator. Thus we need  p \geq 2q  for the 3's to cancel. (The 7's will be cancelled easily as  p \geq q )

Statement 1:

This violates  p \geq 2q , thus we know the 3's will not cancel and the fraction will not terminate. Sufficeint.

Statement 2:

With this, we know that the 7's will cancel but we can't tell if there are enough 3's in the numerator. Insufficient.

Ans: A

SaquibHGMATWhiz · Answer

Solution:      
  Pre Analysis:  
  We are asked if  \frac{(21)^p}{(630)^q}  is terminating or not
 ⇒\frac{(21)^p}{(63	imes 10)^q} 
 ⇒\frac{(3	imes 7)^p}{(3^2	imes 7	imes 10)^q} 
 ⇒\frac{3^{p-2q}	imes 7^{p-q}}{10^q}  
 For the above expression to be terminating  p has to be greater than 2q which will make p greater than q also  
 If a statement is able to answer this with a YES or NO, it is sufficient

Statement 1:   p < 2q
  This statement is directly telling me that p is less than 2q which means  \frac{3^{p-2q}	imes 7^{p-q}}{10^q}  is not terminating 
 Thus, the  statement 1 alone is sufficient and we can eliminate options B, C and E

Statement 2:   p > q
  This is not enough to say if p is greater than 2q or not 
 Thus, statement 2 alone is not sufficient

Hence the right answer is    Option A

TBT · Answer

Can someone give an example of statement 2 which give yes and no as the answer.

bumpbot · Answer

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If p and q are positive integers, is 21^p/630^q a terminating

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If p and q are positive integers, is 21^p/630^q a terminating

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