The statement of the problem
En(p. 436)
Fr(p. 437)
ANSWER CiP
Smallest prime factor 3; largest prime factor 673
Solution CiP
According to the formula
$1+2+\cdot \cdot \cdot +n=\frac{n(n+1)}{2}$
the number is
$M=2(1+2+\cdots +2018)+2019=2018 \cdot 2019+2019=2019^{2}$.
Because 2019 has the decomposition into prime factors $3 \cdot 673$ we get the answer.
$\blacksquare$
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Added May 19, 2021
Good answer see V47n04, pages 176-177
And they get the result $M=2019^2$ and after applying the divisibility with 3 $(2+0+1+9=12=4 \cdot 3 )$ they obtain $2019^2=3^2 \cdot 673^2$. They further show that 637 is a prime number: check the divisibility criterions with 2, 3, 5 and 11; for 7, 13, 17, 19 and 23 (because $25^2=625 < 673< 676=26^2$) apply the division with the remainder ...
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