Question

w, x, y, and z are consecutive positive integers and w<x<y<z.
Column A: The remainder when \((w+x)(x+y)(y+z)\) is divided by 2
Column B: 1

• A. The quantity in Column A is greater.
• B. The quantity in Column B is greater.
• C. The two quantities are equal.
• D. The relationship cannot be determined from the information given.

Hint:

Remember these key parity rules: Even + Odd = Odd, Even \(\times\) Anything = Even, Odd \(\times\) Odd = Odd. The sum of any two consecutive integers is always odd. Since \(w,x\), \(x,y\), and \(y,z\) are all pairs of consecutive integers, each term in the product is odd, making the entire product odd.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This question deals with number properties, specifically the parity (evenness or oddness) of sums and products of consecutive integers. The remainder when a number is divided by 2 tells us if the number is even (remainder 0) or odd (remainder 1).
Step 2: Detailed Explanation:
Consecutive integers always follow an alternating pattern of even and odd. We can test both possible starting cases.
Case 1: The first integer, w, is even.
If \(w\) is even, then the sequence of parities is:
w = Even, x = Odd, y = Even, z = Odd.
Now let's find the parity of the sums in the expression:
\begin{itemize} \item \(w + x\) = Even + Odd = Odd \item \(x + y\) = Odd + Even = Odd \item \(y + z\) = Even + Odd = Odd \end{itemize} The product is \((w+x)(x+y)(y+z)\) = Odd \(\times\) Odd \(\times\) Odd = Odd.
When an odd number is divided by 2, the remainder is 1.
Case 2: The first integer, w, is odd.
If \(w\) is odd, then the sequence of parities is:
w = Odd, x = Even, y = Odd, z = Even.
Now let's find the parity of the sums in the expression:
\begin{itemize} \item \(w + x\) = Odd + Even = Odd \item \(x + y\) = Even + Odd = Odd \item \(y + z\) = Odd + Even = Odd \end{itemize} The product is \((w+x)(x+y)(y+z)\) = Odd \(\times\) Odd \(\times\) Odd = Odd.
Again, when an odd number is divided by 2, the remainder is 1.
Step 3: Final Answer:
In both possible cases, the expression \((w+x)(x+y)(y+z)\) evaluates to an odd number. The remainder when any odd integer is divided by 2 is always 1. Therefore, the quantity in Column A is 1, which is equal to the quantity in Column B.