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

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BSchool Forum Moderator
Joined: 26 Feb 2016
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Concentration: General Management, Leadership
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If p and q are positive integers, is 21^p/630^q a terminating [#permalink]

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31 Dec 2017, 23:02
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Difficulty:

65% (hard)

Question Stats:

43% (01:17) correct 57% (01:18) wrong based on 14 sessions

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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
[Reveal] Spoiler: OA

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Kudos [?]: 830 [0], given: 22

examPAL Representative
Joined: 07 Dec 2017
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Kudos [?]: 23 [0], given: 3

Re: If p and q are positive integers, is 21^p/630^q a terminating [#permalink]

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01 Jan 2018, 01:20
pushpitkc wrote:
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

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.
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Re: If p and q are positive integers, is 21^p/630^q a terminating   [#permalink] 01 Jan 2018, 01:20
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# If p and q are positive integers, is 21^p/630^q a terminating

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