If $w=2^{129} \times 3^{81} \times 5^{128}, x=2^{127} \times 3^{81} \times 5^{128}, y=2^{126} \times 3^{82} \times 5^{128}$, and $z=2^{125} \times 3^{82} \times 5^{129}$, then the order from smallest to largest is
(A) $w, x, y, z$
(B) $x, w, y, z$
(C) $x, y, z, w$
(D) $z, y, x, w$
(E) $x, w, z, y$
If $w=2^{129} \times 3^{81} \times 5^{128}, x=2^{127} \times 3^{81} \times 5^{128}, y=2^{126} \times 3^{82} \times 5^{128}$, and $z=2^{125} \times 3^{82} \times 5^{129}$, then the order from smallest to largest is
(A) $w, x, y, z$
(B) $x, w, y, z$
(C) $x, y, z, w$
(D) $z, y, x, w$
(E) $x, w, z, y$

## Solution

We start with the observation that $2^{125} \times 3^{81} \times 5^{128}$ is a common factor to each of the given numbers.

For the basis of comparison, we remove the common factor and write the numbers as follows:

$$
\begin{aligned}
& w=2^{4} \cdot 5^{3}\left(2^{125} \times 3^{81} \times 5^{128}\right)=2000 k \\
& x=2^{2} \cdot 5^{3}\left(2^{125} \times 3^{81} \times 5^{128}\right)=500 k \\
& y=2 \cdot 3 \cdot 5^{3}\left(2^{125} \times 3^{81} \times 5^{128}\right)=750 k \\
& z=3 \cdot 5^{4}\left(2^{125} \times 3^{81} \times 5^{128}\right)=1875 k, \text { where } k=2^{125} \times 3^{81} \times 5^{128}
\end{aligned}
$$

Thus, $x<y<z<w$.

ANSWER: (C)