Divisibility of products.

Question

Given the products:
1*2*3, 2*3*4..., 2020*2021*2022. How many of these products are divisible by the number 2020?  

To clarify the product is:
p(p+1)(p+2), where p=1, 2, 3... 2020

Answer 1

Easy and obvious solutions are 2018×2019×2020, 2019×2020×2021, 2020×2021×2022.

REVISION FOLLOWING COMMENT

USING QUADRATICS

We need to consider p(p+1)=2020n where n is an integer.

This gives us the quadratic p²+p-2020n=0, so p=(-1±√(8080n+1))/2.

8080n+1=x² where x is an integer. When n=5, x²=40401=201², so p=(-1±201)/2=-101 or 100.

So p(p+1)=100×101=10100. This gives us also p+1=100 and p+2=101, p=99.

Therefore 100×101×102 and 99×100×101 are also divisible by 2020.

When n=81, p=(-1±√654481)/2=404 or -405.

Therefore 404×405×406 and 403×404×405 are also divisible by 2020.

When n=126, p=(-1±√1018081)/2=404 or -405.

Therefore 504×505×506 and 503×504×505 are also divisible by 2020.

FACTOR PAIRS

A better way than using the quadratic is to look at the factors of 2020:

2020=2²×5×101, creating factor pairs (2,1010), (4,505), (5,404), (10,202), (20,101).

If we use quadratics we also have to consider p(p+2)=2020n, p²+2p-2020n=0. So 1010×1011×1012 is also divisible by 2020. As is 200×201×202. These solutions can also be deduced from the factor pairs. That makes 11 in all.

Comments

What about 100*101*102 and 99*100*101?

---

Yes, sorry I missed that, although I did factor 2020=4×5×101 but thought there could be no suitable consecutive multiples. My bad. It's a good question.