If p and q are consecutive even integers and p

Question

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

militarytocorp · Answer

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–pqpq2–pq

Substitute 2 and 4 for even integers, B, C, D can be ruled out.
check again for 4 & 6, only E will suffice the condition.

ans. E

vishumangal · Answer

By substituting p as 2 and q as 4 answer can be find out

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Archit3110 · Answer

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..

lucasd14 · Answer

Hi  Bunuel , is there any way we can validate the correct answer without plugging in numbers? Thanks!

ruccipotato · Answer

What about if we use 0 and 2? All the options, besides A, turn out divisible by 3.

lucasd14 · Answer

If you use 0, then all QA become divisible by 3. Choosing 0 won't be useful.

Bunuel · Answer

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.

zishanjamil94 · Answer

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 * p+2 * p-1 => let's consider p =2 for validation then p+1 will always be 3 , which is divisible by 3.

ScottTargetTestPrep · Answer

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

aryanxx09 · Answer

Yeah, but the question says "must be", so it should ideally work? I think the question should have mentioned positive even integer

If p and q are consecutive even integers and p < q, which of the follo

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GMAT Problem Solving (PS) Questions

If p and q are consecutive even integers and p < q, which of the follo

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