Obviously any pair in the form $(a,1)$ can't be pupusa. But the pair $(391,2)$ is a pupusa one therefore the answer is $b=2$

Note the ineq sign. But (2,2) works beautifully.

You mean $(4,2)$, right ?. Update: I read it gcd, so I now think the answer is $(1,392)$

jg123 wrote: Note the ineq sign. But (2,2) works beautifully. Kayak wrote: You mean $(4,2)$, right ? But $lcm(391,2)=782>2=lcm(2,2)$ and $lcm(391,4)=1564>4=lcm(4,2)$ Anyway,I think I have seen the similar question with the condition $a<391$ before.

NotABot wrote: We call a pair $(a,b)$ of positive integers pupusa if $$\textup{lcm}(a,b)>\textup{lcm}(a,391)$$ Find the minimum value of $b$ across all pupusa pairs. Fun Fact: OMCC 2017 was held in El Salvador. Pupusa is their national dish. It is a corn tortilla filled with cheese, meat, etc. This was not the statement of the problem. The English version is available at: day1, day2

The OP forgot to add that $a<391$. Note that $391=17\cdot23$ We have that $lcm(a,391)$ is one of $\{a, 17a, 23a, 391a\}$. Since $a<391$ we can't have $lcm(a,391)=a$, so $lcm(a,391)\geq 17a$. Therefore $b\geq18$ (else $lcm(a,b)\leq ab\leq17a$). The pair $(23,18)$ is pupusa since $lcm(23, 391)=391<23\cdot18=lcm(23,18)$ so the minimum value of $b$ is $18$.

First, note that, $391 = 17.23$, so $gcd(a,391) = 1, 17 or 23$, because $"a" < 391$, denote by $gcd(a,391) = d$, and $gcd(a, b) = e$, then $lcm(a,391) = 391a/d.$ Further $lcm(a, b) = ab/e$, thus, $ab/e > 391a/d$, implying $b/e > 391/d$, and $b > 391e/d$, from this we can maximize $"d"$ and minimize $"e"$, for this we only need to make $"a"$ and $"b"$ coprimes$(e = 1)$, and lastly we can take $"a"$ such that $gcd(a, 391) = d = 23$, the maximum possible value. Therefore, from $b > 391e/d$ we conclude that $b > 391.1/23 = 17$, so $b >= 18$, one example of pupusa pair for $"b" = 18$ is $"a" = 23$, because $d = 23$ and $e = 1$. Finishing, $18$ is the minimum possible value for b! Q.E.D.

\begin{align} [a,b]&>[a,391] \nonumber \\ [a,b](a,b)(a,391)&>(a,b)(a,391)[a,391] \nonumber\\ ab(a,391)&>a\cdot391(a,b)\nonumber\\ b&>\frac{391(a,b)}{(a,391)}\nonumber \end{align}But, $\frac{391(a,b)}{(a,391)}\geq\frac{391}{(a,391)}\geq\frac{391}{23}=17$, then $b\geq 18$. $a=23$,$b=18$ works. So the minimum possible value for $b$ is 18.

We claim that $18$ is the minimum value of $b$. A valid construction is $a=23,b=18$. Now we try to prove it. Notice that the prime factorization of $391$ is $17\times 23$. Hence $lcm(a,391)\geq 17a$ and $a\mid lcm(a,391), lcm(a,b)$. Therefore, $lcm(a,b)\geq 18a$ and so $b\geq 18$. Hence the answer is $18$.