A number theory problem by أحمد الحلاق

After analysing the differences between consecutive terms and their differences, we can option a formula for the quadratic sequence:

$q_n = n^2 + n = n × (n + 1)$

We can also find, that:

2 = 1 × 2 and 14520 = 120 × 121

Therefore, 1 ≤ n ≤ 120.

Now, if we factorise our divisor, we get:

$120 = 2^3 × 3 × 5 = 8 × 3 × 5$

Since n and (n +1) are consecutive integers, exactly one of them will be even and the other will be odd. This also means, that exactly one of them has to be divisible by 8.

So we are looking for pairs of consecutive numbers (n , n+1), where: 120 | n(n+1).

There are 15 integers within the (1 , 121) interval, which are divisible by 8:

$k \in (8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120)$

Now, we are looking for those (k - 1, k) and (k, k+1) pairs, where one of the integers in the pair is divisible by 3 and the other by 5 (or one of them is divisible by both 3 and 5).

We can find 8 of such pairs:

(15, 16); (24, 25); (39, 40); (80, 81); (95, 96); (104, 105); (119, 120) and (120, 121).

$\text {Hence, our answer is } \boxed {8}$