"Sequence of 2's" S is the infinite sequence S1 = 2, ...

[Nemps7]

S is the infinite sequence S1 = 2, S2 = 22, S3 = 222,...Sk = Sk-1 + 2(10k-1). If p is the sum of the first 30 terms of S, what is the eleventh digit of p, counting right to left from the units digit?

Answer choices are
A: 1
B: 2
C: 4
D: 6
E: 9

What I'm having trouble with is understanding the MGMAT explanation for the shortcut, or trick to the problem.

I've copied it below:

We could also have seen that each column has one less 2 than the previous, so if we started out with 30 2’s in the first column, the 11th column must have 11 - 1 = 10 less 2’s, for a total of 20 2’s. The amount that is carried over from the previous column could be calculated by realizing that the 10th column had 21 2’s for a total of 42. Since there is no way that the 10th column inherited more than 8 from the 9th column, the total must be forty-something and the amount that is carried over to the 11th column MUST BE 4. This makes the total for the 11th column 40 + 4 = 44 and the 11th digit of p 4.

The correct answer is C.

MY QUESTION: Where I get lost is "Since there is no way that the 10th column inhereted more than 8..." I imagine this has something to do with the 42, but I can't for the life of me figure out why it should be clear that the 10th column can't inherit more than 8?

Can anyone clarify why this should be so obvious, and why I fail to see it?

I understand the total must be 40 + something carried from the previous column, but I can't understand how we arrive at 4.

Thank you for the help!
Ned

[RonPurewal]

why it should be clear that the 10th column can't inherit more than 8?

That would require 80 or more in the next column to the right. To get that, you'd need to add waaaaayyyyy more 2's than you have available.

[SowmyaM788]

It helps to jot it down on paper - a sequence of 2's one below the other, with the rightmost column having 30 x 2's, right but one having 29 x 2's and so on..
Now the 10th column will have 21 x 2's and the 11th has 20 x 2's.
So the 11th adds up to a value of 40. We now want to know how much was carried over from the 10th to 11th column.
We can safely say the sum of the 9th column will not exceed 80. This is because the actual sum of 9th column is 22x2= 44 + a carry over. This sum cannot be greater than 80 for sure.
Since the carry over from the previous column is the TENS digit of the sum of the previous column, sum of the 10th column will still be 42 + (carryover < 8 from 9th column) = 40 something (less than 50)
So 4 gets carried over to the 11th column.
So sum of 11th column = 40 + 4 = 44.

[Sage Pearce-Higgins]

Good logic, SowmyaM788.