100,000ths place. You are welcome
20 seconds faster than Jesse so one minute and 57 seconds is my best guess.
Might have to experiment a bit to choose the right answer.
In A, the first term is 456 and the common difference is 10. Each time we have a new term, the next one is the same except that 10 is added.
Suppose n were 1000. Then we'd have 456 + (1000)(10) = 10456
In B, the first term is 5 and the common ratio is 3. From 5 we get 15 by mult. 5 by 3. Similarly, from 135 we get 405 by mult. 135 by 3. This is a geom. series with first term 5 and common ratio 3. a_n = a_0*(3)^(n-1).
So if n were to reach 1000, the 1000th term would be 5*3^999, which is a very large number, certainly more than the 10456 you'd reach in A, above.
Can you now examine C and D in the same manner, and then choose the greatest final value? Safe to continue using n = 1000.
The number of 10 chips stacks that Dave can make if two stacks are not considered distinct is 110.
The solution
To get the symmetric stacks, one has to subtract the symmetric stacks to know the ones that are asymmetric.
The symmetric are flipped. Given that they are double counted what we have to do is to divide through by 2.
6/2 = 3
10/2 = 5
4/2 = 2
1/2(10C6) - (5C3) + (5C3)
0.5(210-10+10)
= 110
