A number theory problem by Viki Zeta

Every number in N is going to have a factor of 6! = 720, therefore N / 720 will have a quotient with remainder 0.

$N = \sum_{i=20}^{120} i!\\ N = 20! + 21! + 22! + \ldots + 120! \\ \text{We know that, 6! = 720 and also} \\ N = (6! \times 7 \times \ldots \times 20) + (6! \times 7 \times \ldots \times 21) + \ldots + (6! \times 7 \times \ldots \times 120) \\ N = 6! ( (7 \times \ldots \times 20) + (7 \times \ldots \times 20) + \ldots + (7 \times \ldots \times 20) ) \\ N = 6! \times m \text{, m = }( (7 \times \ldots \times 20) + (7 \times \ldots \times 20) + \ldots + (7 \times \ldots \times 20) ) \\ N = 6! \times m + 0 \\ \text{Now, using Euclids division lemma, number N, When divided by 6!, leaves remainder `0'} \\ \therefore \text{The remainder is } 0$