Solution adb1904a-f892-4867-9ae8-07f8cfb58f9a

Points: 1 pt

We see
\[
60=(2^2)(3)(5),
\]
and the divisors are
\[
\{1,\,2,\,3,\,2^2,\,5,\,(2)(3),\,(2)(5),\,(2^2)(3),\,
(3)(5),\,(2^2)(5),\,(2)(3)(5),\,(2^2)(3)(5)\}.
\]

However, using
\[
1=2^0=3^0=5^0,
\]
we can rewrite all the numbers as
\[
\left\{
\begin{array}{cccc}
2^0 3^0 5^0 & 2^0 3^1 5^0 & 2^0 3^0 5^1 & 2^0 3^1 5^1 \\
2^1 3^0 5^0 & 2^1 3^1 5^0 & 2^1 3^0 5^1 & 2^1 3^1 5^1 \\
2^2 3^0 5^0 & 2^2 3^1 5^0 & 2^2 3^0 5^1 & 2^2 3^1 5^1
\end{array}
\right\}.
\]

We see that every exponent for a prime is less than or equal to the corresponding exponent in $60$.