Q, R, S, and T are four consecutive positive integers, where Q

Question

Q, R, S, and T are four consecutive positive integers, where Q < R < S < T. Which of the following expressions must be odd?

A. QR + ST.
B. Q + R + S + T.
C. Q^2 + S^3.
D. Q^2 + R^2.
E. Q^2 + 2R

Mohammadmo · Answer

As these four numbers are positive integers so, none of them is zero.
And because these are consecutive so, both Q and S and both R and T are the same in terms of being odd or even.
Hence, Q²+R²=odd number.
Because certainly, one of them is odd.
Option D

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sj296 · Answer

Q < R < S < T

Just take options one by one:
A) QR + ST
Let's say Q is odd then QR = Even (R will be even if Q is odd)
Similarly, ST = Even

Hence, Even + Even = Even
Therefore, option A is wrong.

B) Q+R+S+T
2 odd and 2 even
=> Odd + Even + Odd + Even = Even
=> Even + Odd + Even + Odd = Even
Therefore, option B is wrong

C) Q^2 + S^3
Let's say Q = odd then S will be Odd
=> Q^2 = Odd and S^3 = Odd => Odd + Odd = Even
If Q is even then S will be Even => Q^2 + S^3 = Even
Therefore, option C is wrong

D) Q^2 + R^2
Assuming Q = Odd then R = Even
=> Odd + Even = Odd
Assuming Q = Even then R = Odd
=> Even + Odd = Odd
Therefore, option D is correct option

E) Q^2 + 2RD
If Q is even then Q^2 = even
=> Even + Even = Even
If Q is odd then Q^2 = Odd
=> Odd + Even = Odd
Not always odd therefore, option E is also wrong

ScottTargetTestPrep · Answer

Consecutive integers always have the pattern even-odd-even-odd (or odd-even-odd-even). Thus we say that two consecutive integers have opposite parity i.e., one even one odd.

Since Q and R will always have opposite parity - meaning one is odd and the other is even or vice versa- Q^2 + R^2 will always be odd since the square of one of them will be even and the square of the other will be odd, and even + odd = odd.

Answer: D

Q, R, S, and T are four consecutive positive integers, where Q < R < S

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