Question

$x, y, z$ are three positive integers such that $x>y>z$. Which of the following is closest to the product $xyz$?

• A. $(x - 1)yz$
• B. $x(y - 1)z$
• C. $xy(z - 1)$
• D. $x(y + 1)z$

Hint:

When approximating expressions, plug in small integer values that satisfy the condition and compare outputs.

Answer 1

The Correct Option is A

Let’s try values: $x = 6$, $y = 5$, $z = 4$ \[ xyz = 6 \cdot 5 \cdot 4 = 120 \] Now test each option: - (a) $(x-1)yz = 5 \cdot 5 \cdot 4 = 100$ - (b) $x(y-1)z = 6 \cdot 4 \cdot 4 = 96$ - (c) $xy(z-1) = 6 \cdot 5 \cdot 3 = 90$ - (d) $x(y+1)z = 6 \cdot 6 \cdot 4 = 144$ Closest to 120 = $\boxed{100}$ from option (a) \[ \boxed{(x - 1)yz} \]