# The math problem only 1% of fb users can solve

You know the kind, right? The math problems only 1 percent of the worlds population can solve. These constantly pop up on fb. As the above. The 1.5 million commentators are probably in that 1 percentile. Maybe they aren't. Facebook has close to $$3 \cdot 10^{9}$$ user (source). Which is what, 1/3 of the world population. So does the number of people able to solve, scale? Is it 1% of fb users who are able to solve? (maybe those users able to are busy doing other things) Are anyone? Who cares?

Anyway, I think this problem is kind of interesting. First of all the solution is $$n_i = i \cdot (i + 1)$$

We can easily print up till $$i = 9$$ in python:

[i=0] 0*1 = 0 [i=1] 1*2 = 2 [i=2] 2*3 = 6 [i=3] 3*4 = 12 [i=4] 4*5 = 20 [i=5] 5*6 = 30 [i=6] 6*7 = 42 [i=7] 7*8 = 56 [i=8] 8*9 = 72 [i=9] 9*10 = 90

However we can also print the difference between $$n_i$$ and $$n_{i + 1}$$. We get

n[0+1]-n[0] = 2-0 = 2 n[1+1]-n[1] = 6-2 = 4 n[2+1]-n[2] = 12-6 = 6 n[3+1]-n[3] = 20-12 = 8 n[4+1]-n[4] = 30-20 = 10 n[5+1]-n[5] = 42-30 = 12 n[6+1]-n[6] = 56-42 = 14 n[7+1]-n[7] = 72-56 = 16 n[8+1]-n[8] = 90-72 = 18

The differences contains one common factor. For example we have $$4 \cdot 5 - 3 \cdot 4 = 8$$ Since ring (distribute) we can rewrite $$4(5 - 3) = 4 \cdot 2 = 8$$ This common factor gets 1 bigger for every next $$i$$. The difference between the non-common factors is always $$2$$. That is why we have the differences as the list $$[2,4,6,8,10,...]$$. We can formulate the difference, we get: $$n_{i + 1} - n_i = (i + 1)(i + 2) - i (i + 1)$$ we do things, and $$n_{i + 1} - n_i = (i + 1)(i + 2 - i)$$ we do more things, and $$n_{i + 1} - n_i = 2(i + 1)$$ So for $$i = 3$$ we have $$n_i = 3 \cdot (3 + 1) = 12$$ and for $$i = 4$$ we have $$n_i = 4 \cdot (4 + 1) = 20$$ and $$n_4 - n_3 = 20 - 12 = 2(3 + 1) = 8$$ I can't think of more to say about this problem.