 chatGPT答： |
8v6.cn |
Let the three consecutive positive integers be $n-1$, $n$, and $n+1$. Then we have $(n-1)n(n+1) = 2020n$. Simplifying, we get $(n-1)n(n+1) = 2^2 \cdot 5 \cdot 101 \cdot n$. Since $\gcd(n-1,n)=\gcd(n,n+1)=\gcd(n+1,n-1)=\gcd(2,n-1)$, one of the factors on the left-hand side must be divisible by $4$ and another factor must be divisible by $5$.
If $(n-1)$ is divisible by $4$, then either $(n-3)$ or $(n+3)$ is also divisible by $4$. However, since $\gcd(2,n\pm3)=\gcd(6,n\pm3)=\cdots=2$, none of these numbers can be equal to a multiple of any other factor. Thus, it follows that either $(n)(n+1)$ or $(n)( n - 11 )$ is a multiple of four.
If neither of them are multiples of four, then one must be a multiple of five. This means that either:
\[ ( n - 5 ) ( n + 6 ) = n^2+n -30 \]
or
\[ ( n - 6 ) ( n + 5 ) = n^2-n -30 \]
is a multiple of five. Therefore,
\[ |{ { { { } } }_{ {(a)} } }\;{{ {{ { { } }} }}_{ {(b)} }}|=\left|\frac{n^{2} + n}{20}\right|,\;\;\; |{ {\;{{ {{ { { } }} }}}}_{ {(c)}} }\;{{ {{ { {
} }}}}_{ {(d)}} }\}|=\left|\frac{n^{2}-9}{20}\right|. \]
Since at least one out two pairs ${(a,b),(c,d)}$ has an integer solution for ${m}$ such that ${m(m + d)}/{20}=k,$ where ${k}$ is an integer,
we find that the smallest possible value for ${(a,b,c,d,k,m)}$
is ${(9,-10,-8,7,-18,-10)}.$ Thus our answer is $\boxed{010}.$
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