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

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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. \(pq^2-pq\)

Answer: E

Source: Princeton Review

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M7MBA wrote:
Thu Nov 19, 2020 9:03 am
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. \(pq^2-pq\)

Answer: E

Solution:

We can let p = 2 and q = 4 and check the answer choices given.

A. p^2 + pq = 2^2 + 2(4) = 4 + 8 = 12 → This is divisible by 3.

B. p(q^2) + pq = 2(4)^2 + 2(4) = 32 + 8 = 40 → This is not divisible by 3.

C. (p^2)q - pq = 2^2(4) - 2(4) = 16 - 8 = 8 → This is not divisible by 3.

D. p^2q^2 = 2^2(4^2) = 4(16) = 64 → This is not divisible by 3.

E. p(q^2) - pq = 2(4)^2 - 2(4) = 32 - 8 = 24 → This is divisible by 3.

We see that both A and E could be the correct answer when p = 2 and q = 4. To determine which is the correct answer, we can let p = 4 and q = 6 and check only choices A and E.

A. p^2 + pq = 4^2 + 4(6) = 16 + 24 = 40 → This is not divisible by 3.

By default, choice E will be the correct answer. However, let’s check it to make sure.

E. p(q^2) - pq = 4(6)^2 - 4(6) = 144 - 24 = 120 → This is divisible by 3.

Answer: E

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