(To reflect a puzzle that came up while walking,
here’s a general solution that also came up while walking.
This write-up was reorganized a few days after posting.)
$\begingroup
\def \b #1{ \boldsymbol{#1} }
\def \F #1#2{ {\LARGE\strut} \f{#1}{#2} }
\def \f #1#2{ { \large #1 \over \large #2 } }
\def \Q #1#2{ { \large\frac{2^{\Large\kern.05em\b{#1}}}{#2} } }
\def \t #1{ {\small\textsf{#1}} }
\def \x { { \scriptsize\raise.2ex\times}\kern.1em }
\def \& #1{ & \kern-.9em #1 \kern-1em & }
\def \+ { { \kern.1em + } }
\def \/ { \\[-.4ex] }
$The probability that $n(n\+1)$ is a multiple of $100$ is ...
$$ { 2^{\large (\textsf{how many distinct prime factors of}~100)} \over 100 } ~=~ { 2^{\large 2} \over 100} ~=~ { 1 \over 25} $$
This works in general for $n(n\+1)$
that might be a multiple of any $m \ge 2$.
(In the puzzle, $m=100$.)
$$ \begin{matrix}
\begin{matrix} \\ m \\ \t{(as an example)} \end{matrix}
&& \begin{matrix} \\ \t{Prime} \/ \t{factorization} \/ \t{of}~ m \end{matrix}
&& \begin{matrix} \strut D(m) ~ = \/ \t{how many} \/ \t{distinct prime}
\/ \t{factors of}~ m \end{matrix}
&& \begin{matrix} {\Large{2^{D(m)} \over \raise.3ex m}} ~ =
\/ \t{probability that}
\/ n(n\+1) ~\t{is a}
\/ \t{multiple of}~ m \end{matrix}
\\[1ex]\hline 2 && 2^1 && 1 && \Q{1}{2} ~=~ 1 \\
3 && 3^1 && 1 && \Q{1}{3} ~=~ \F{2}{3} \\
4 && 2^2 && 1 && \Q{1}{4} ~=~ \F{1}{2} \\
6 && 2^1 \x 3^1 && 2 && \Q{2}{6} ~=~ \F{2}{3} \\
72 && 2^3 \x 3^2 && 2 && \Q{2}{72} ~=~ \F{1}{18 } \\
\boldsymbol{100}&&\b2^2 \x \b5^2 &&\b2 && \t{(already mentioned)} \\
4725 && 3^3 \x 5^2 \x 7^1 && 3 && \Q{3}{4725} ~=~ \F{8}{4725}
\end{matrix}$$
Sure enough, the probabilities are the same
for $m = 3$ and $m = 6 = 3 \x 2$.
After all, $n(n\+1)$ is always divisible by $2$.
And, sure enough,
$ 2^{\large (\textsf{how many distinct prime factors of}~4725)} {=}~ 8 $
instances of $n(n\+1)$ among the $4725$ values of
$1 \le n \le 4725$ are divisible by $m = 4725 = 3^3 \x 5^2 \x 7^1 $.
$$\small\begin{array}{rrlcrrlrrrlcrrl}
350 \&\x 351 &=& ( 5^2 \x 7 ) \&\x 2 ~\&\x~ 13 \&\x ( 3^3 ) &=& 26 \&\x 4725 \\
1350 \&\x 1351 &=& ( 3^3 \x 5^2 ) \&\x 2 ~\&\x~ 193 \&\x ( 7 ) &=& 386 \&\x 4725 \\
1700 \&\x 1701 &=& ( 5^2 ) \&\x 68 ~\&\x~ 9 \&\x ( 3^3 \x 7 ) &=& 612 \&\x 4725 \\
3024 \&\x 3025 &=& ( 3^3 \x 7 ) \&\x 16 ~\&\x~ 121 \&\x ( 5^2 ) &=& 1936 \&\x 4725 \\
3374 \&\x 3375 &=& ( 7 ) \&\x 482 ~\&\x~ 5 \&\x ( 3^3 \x 5^2 ) &=& 2410 \&\x 4725 \\
4374 \&\x 4375 &=& ( 3^3 ) \&\x 162 ~\&\x~ 25 \&\x ( 5^2 \x 7 ) &=& 4050 \&\x 4725 \\
4724 \&\x 4725 &=& \&{} 4724 ~\&\x~ 1 \&\x ( 3^3 \x 5^2 \x 7 ) &=& 4724 \&\x 4725 \\
4725 \&\x 4726 &=& ( 3^3 \x 5^2 \x 7 ) \&\x 1 ~\&\x~ 4726 \&{} &=& 4726 \&\x 4725
\end{array}$$
Why this works, with $m = 4725$ as an example:
The combined prime factors of $n$ and $n\+1$
must include the prime factors of $4725$,
namely at least three $3$s, two $5$s and one $7$.
All $3$s may be factors only of $n$ or only of $n\+1$
as consecutive numbers have no common factors.
Same for the $5$s and $7$, so
$27$ ($\small =3^3 \raise.7ex\strut$),
$25$ ($\small =5^2 \raise.7ex\strut$)
and $7$ are, in effect, indecomposable factors.
The possibilities of $27$, $25$ or $7$ being factors of a number
are independent of each other because $27$, $25$ and $7$
share no common factors.
The probability that $27$ is a factor of $n$ or $n\+1$ is $\f{2}{27}$.
The probability that $25$ is a factor of $n$ or $n\+1$ is $\f{2}{25}$.
The probability that $7$ is a factor of $n$ or $n\+1$ is $\f{2}{7}$.
As these component probabilities are independent,
multiply them together for the probability that
$27$, $25$ and $7$ are all present as factors of $n$($n\+1)$:
$$ \small {2 \over 27} \x {2 \over 25} \x {2 \over 7}
= {2^3 \over 27 \x 25 \x 7} = {8 \over 4725} $$
In general, the resulting probability fraction has:
Numerator equal to 2 raised to the power that is the count
of distinct prime factors of the number whose multiples are sought.
(In the puzzle, $m = 100 = \b2^2 \x \b5^2 \raise.3ex\strut$
has $2$ distinct prime factors.)
This is because each distinct prime factor
may appear in either $n$ or $n\+1$.
Denominator equal to the number ($m=100$ in the puzzle)
whose multiples are sought.
This is because the denominators of the multiplied component probabilities
are collectively the prime factors of $m$ raised to their exponents.
$\endgroup$