we count units, tens, hundreds, etc. - all 2s (two-s)
p is the sum of the first 30 terms means we have 30 units, 29 tens, 28 hundreds, etc. ... and all units, tens, hundreds, and so on have the value of 2

counting right to left we have tens billions value (digit)

30*2=60
29*20=580
28*200=5600
27*2000=54000
...
pattern ===> (30-n)*2~n zeros, makes approximately (30-9)*2~9 zeros ===> 21*2,000,000,000

so, 60+580+5600+54000+(26*20000, or 520000)+ ... + 42 bln.. The number before 42 bln was 22*200,000,000 (or 4,400,000,000) would add to 46 bln and the other previous numbers would add to less than tens of billions, hence 4.

nafiul9090 wrote:S is the infinite sequence S1 = 2, S2 = 22, S3 = 222,...Sk = S^(k-1) + 2((10^(k-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?

A) 1
B) 2
C) 4
D) 5
E) 9

any short cut method to solve this type of problem....