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Re: If p and q are consecutive even integers and p < q, which of the follo [#permalink]
If p and q are consecutive even integers and p < q, which of the following must be divisible by 3 ?


A. p2+pqp2+pq

B. pq2+pqpq2+pq

C. p2q–pqp2q–pq

D. p2q2p2q2

E. pq2–pq

Upon checking with values of p =2 and q=4 we get option E to satisfy the relation to proof..
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Re: If p and q are consecutive even integers and p < q, which of the follo [#permalink]
Bunuel wrote:
If p and q are consecutive even integers and p < q, which of the following must be divisible by 3 ?


A. \(p^2 + pq\)

B. \(pq^2 + pq\)

C. \(p^2q – pq\)

D. \(p^2q^2\)

E. \(pq^2 – pq\)


Hi Bunuel, is there any way we can validate the correct answer without plugging in numbers? Thanks!
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Re: If p and q are consecutive even integers and p < q, which of the follo [#permalink]
What about if we use 0 and 2? All the options, besides A, turn out divisible by 3.
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Re: If p and q are consecutive even integers and p < q, which of the follo [#permalink]
ruccipotato wrote:
What about if we use 0 and 2? All the options, besides A, turn out divisible by 3.


If you use 0, then all QA become divisible by 3. Choosing 0 won't be useful.
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Re: If p and q are consecutive even integers and p < q, which of the follo [#permalink]
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lucasd14 wrote:
Hi Bunuel, is there any way we can validate the correct answer without plugging in numbers? Thanks!


If p and q are consecutive even integers and p < q, which of the following must be divisible by 3 ?

A. \(p^2 + pq\)

B. \(pq^2 + pq\)

C. \(p^2q – pq\)

D. \(p^2q^2\)

E. \(pq^2 – pq\)

p and q are consecutive even integers and p < q: \(p = n\) and \(q = n + 2\), where \(n\) is an even integer.

A. \(p^2 + pq=(n)^2+(n)(n+2)=2n(n + 1)\). This one will not be divisible by 3 if \(n+2\) is a multiple of 3 (in this case neither \(n\) nor \(n+1\) will be multiples of 3, therefore the product also will not be a multiple of 3. For example, consider \(n+2=0\)).

B. \(pq^2 + pq=(n)(n+2)^2+(n)(n+2)=n(n + 2)(n + 3)\). This one will not be divisible by 3 if \(n+1\) is a multiple of 3 (in this case n, n+2, and n+3 will not be multiples of 3, therefore the product also will not be a multiple of 3. For example, consider \(n+1=0\)).

C. \(p^2q – pq=n^2(n+2)-n(n+2)=(n - 1)n (n + 2) \). This one will not be divisible by 3 if \(n+1\) is a multiple of 3 (in this case \(n-1\), \(n\), and \(n+2\) will not be multiples of 3, therefore the product also will not be a multiple of 3. For example, consider \(n+1=0\)).

D. \(p^2q^2=n^2(n+2)^2=n^2(n + 2)^2\). This one will not be divisible by 3 if \(n+1\) is a multiple of 3 (in this case neither \(n\) nor \(n+2\) will be multiples of 3, therefore the product also will not be a multiple of 3. For example, consider \(n+1=0\)).

E. \(pq^2 – pq=(n)(n+2)^2-(n)(n+2)=n(n + 1)(n + 2) \). This is a product of three consecutive even integers, so one of them will for sure be a multiple of 3, therefore the product will also be a multiple of 3.

Answer E.

But number plugging will be much faster.
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Re: If p and q are consecutive even integers and p < q, which of the follo [#permalink]
an easy way would be cross checking all the options
For option e) p*q^2–p*q
we know q = p+2 (Since q is greater and next consecutive even number)
then p (p+2)^2 - p (p+2)
let's take p+2 common then p*(p+2) [(p+2) - 1] = p * p+2 * p-1 => let's consider p =2 for validation then p+1 will always be 3 , which is divisible by 3.
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Re: If p and q are consecutive even integers and p < q, which of the follo [#permalink]
Expert Reply
Bunuel wrote:
If p and q are consecutive even integers and p < q, which of the following must be divisible by 3 ?


A. \(p^2 + pq\)

B. \(pq^2 + pq\)

C. \(p^2q – pq\)

D. \(p^2q^2\)

E. \(pq^2 – pq\)


Since p and q are consecutive even integers and since p < q, it must be true that q = p + 2. Let's evaluate each answer choice with this piece of information in mind:

Answer choice A: p^2 + pq

\(\Rightarrow\) p^2 + pq

\(\Rightarrow\) p(p + q)

\(\Rightarrow\) p(p + p + 2)

\(\Rightarrow\) p(2p + 4)

\(\Rightarrow\) 2p(p + 2)

This expression is not necessarily divisible by 3, for instance, if p = 2.

Answer choice B: pq^2 + pq

\(\Rightarrow\) pq^2 + pq

\(\Rightarrow\) pq(q + 1)

\(\Rightarrow\) p(p + 2)(p + 2 + 1)

\(\Rightarrow\) p(p + 2)(p + 3)

This expression is not necessarily divisible by 3, for instance, if p = 2.

Answer choice C: (p^2)q – pq

\(\Rightarrow\) (p^2)q - pq

\(\Rightarrow\) pq(p - 1)

\(\Rightarrow\) p(p + 2)(p - 1)

This expression is not necessarily divisible by 3, for instance, if p = 2.

Answer choice D: (p^2)(q^2)

\(\Rightarrow\) (p^2)(q^2)

\(\Rightarrow\) (p^2)(p + 2)^2

Once again, we can let p = 2 to obtain a number which is not divisible by 3. Thus, this expression is not divisible by 3 either.

Since we eliminated all answer choices besides E, the correct answer must be E. However, as an exercise, let's verify that the expression given in answer choice E must be divisible by 3.

Answer choice E: pq^2 - pq

\(\Rightarrow\) pq^2 - pq

\(\Rightarrow\) pq(q - 1)

\(\Rightarrow\) p(p + 2)(p + 2 - 1)

\(\Rightarrow\) p(p + 2)(p + 1)

We see that this expression is the product of three consecutive integers. Since the product of n consecutive integers is always divisible by n!, this expression is divisible by 3 for every value of p.

Answer: E
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Re: If p and q are consecutive even integers and p < q, which of the follo [#permalink]
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Re: If p and q are consecutive even integers and p < q, which of the follo [#permalink]
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