mn2010 wrote:
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?
A) 1
B) 2
C) 4
D) 6
E) 9
Anyone knows what is the easiest and fastest method of solving this ?
--------------------------------------------------2
-------------------------------------------------22
-----------------------------------------------222
---------------------------------------------2,222
-------------------------------------------22,222
...
222,222,222,222,222,222,222,222,222,222
Total 30 numbers.
For the first digit (units place) we should add 30 2's --> 30*2=60, so 0 will be units digit and 6 will be carried over;
For the second digit (tens place) we should add 29 2's --> 29*2=58+6=64, so 4 will be written for this digit and 6 will be carried over;
...
For the 10th digit we should add 21 2's --> 21*2=42, so min value for the number carried over is 4. Max value is also 4, because even if the carry remained 6, as we had at the beginning, still --> 42+6=48, so still 4 will be carried over;
For the 11th digit we should add 20 2's --> 20*2+4=44, so 11th digit will be 4.
Answer: C.
Just curios: is it GMAT question?
Another more risky solution would be to try and find a pattern..
we could assume that this pattern will go on, therefore 11/3 gives as a remainder of 2 therefore the 11th digit will be the same wit the 2nd ---> C