Observed that 7 is a divisor of 14, 21 and 28.
So in effect:
![\bold{14 = 7 * 2}](https://img.qammunity.org/2023/formulas/mathematics/college/le41osss695vogspjhvrkndibb1swiiy6y.png)
![\bold{21 = 7 * 3}](https://img.qammunity.org/2023/formulas/mathematics/college/eg7v5g9n5fn6xvxop5v6larmtt5zfsod70.png)
![\bold{28 = 7 * 4}](https://img.qammunity.org/2023/formulas/mathematics/college/kyzlriot0qcsuk6bybnj7d2drbo2jip7kz.png)
Thus, adding member by member the previous equalities, we would have:
![\bold{14 + 21 + 28 = 63 = 7 * 2 + 7 * 3 + 7 * 4}](https://img.qammunity.org/2023/formulas/mathematics/college/1zews3comdu77h55890g1izsily236mfk0.png)
Taking out the common factor of 7 in the previous expression, we will have:
![\bold{4 + 21 + 28 = 7 * 9}](https://img.qammunity.org/2023/formulas/mathematics/college/d2otykbyd0ksg2k973cy7qco5d98tbezai.png)
I mean:
![\bold{14 + 21 + 28 = 7 * 9}](https://img.qammunity.org/2023/formulas/mathematics/college/vbbx30vdvnn2mbim42ygn18nyiz6xl93ei.png)
That is to say, that 7 is a divisor of the sum
since it is contained exactly 9 times in said sum.