PROBLEM MA95 - Crux Mathematicorum V46N9

      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.

$\blacksquare$ 

=======================================================================

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 ...

=end added=