Solution a420b0bc-47b6-436f-9988-883460302226

Points: 2 pts

Lemma 1

For every $i$, we have
\[
p_i^{a_i}\mid p_i^{\max(a_i,b_i)}
\qquad\text{and}\qquad
p_i^{b_i}\mid p_i^{\max(a_i,b_i)}.
\]
Since $\max(a_i,b_i)\geq a_i$, the exponent in
\[
p_i^{\max(a_i,b_i)-a_i}
\]
is nonnegative, so this number is an integer. Thus
\[
p_i^{\max(a_i,b_i)}
= p_i^{a_i}\,p_i^{\max(a_i,b_i)-a_i}
\]
is an integer multiple of $p_i^{a_i}$. The same argument applies to $b_i$.

Lemma 2: The product is the least common multiple.

Let
\[
L=p_1^{\max(a_1,b_1)}p_2^{\max(a_2,b_2)}\cdots p_r^{\max(a_r,b_r)}.
\]
Multiplying the divisibility statements from Lemma 1 gives
\[
n\mid L\qquad\text{and}\qquad m\mid L.
\]
Now let $c$ be any positive common multiple of $n$ and $m$, and let $c_i$ be the exponent of $p_i$ in the prime factorization of $c$. Since $n\mid c$ and $m\mid c$, unique prime factorization gives
\[
c_i\geq a_i\qquad\text{and}\qquad c_i\geq b_i.
\]
Hence
\[
c_i\geq\max(a_i,b_i)
\quad\Longrightarrow\quad
p_i^{\max(a_i,b_i)}\mid p_i^{c_i}.
\]
Multiplying these statements gives
\[
L\mid p_1^{c_1}p_2^{c_2}\cdots p_r^{c_r}\mid c,
\]
so $L\leq c$. Therefore $L$ is the least common multiple of $n$ and $m$.

Final answer:
\operatorname{lcm}(n,m)=p_1^{\max(a_1,b_1)}p_2^{\max(a_2,b_2)}\cdots p_r^{\max(a_r,b_r)}