S is the infinite sequence S1 = 2, S2 = 22, S3 = 222,...Sk = Sk1 + 2(

Hey Bunuel,

Is it possible to come up with a sequence formula for this question.
Because we have the value (2) of first term and with the sequnece formula we can get the value of the last term (30th term).

Once we have the first and last term values, we can take the average of the first and last values and multiply that by 30 to get the sum of 30 terms. That way we can get the eleventh digit of p.

Please share your thoughts on the same.

Formula Sum = average of the first and last terms multiplied by # of terms can be used for evenly spaced set, but 2, 22, 222, ... is not such set, hence you can not use this formula here.

Thanks Bunuel. Makes sense. I forgot about the evenly spaced set and hence mixed it up.

Another more risky solution would be to try and find a pattern..

1st digit ---> 30*2=60 ---> 0 and 6 will be carried over
2nd digit ---> 29*2+6=64 ---> 4 and 6 will be carried over
3rd digit ---> 28*2+6=62 ----> 2 and 6 will be carried over
4th digit --->27*2+6=60 ----> 0 and 6 will be carried over

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

is my reasoning correct bunuel?

Unfortunately not.
If you continue you'll see that there is no pattern of 3:
5th digit ---> 26*2+6=58 ---> 8 and 5 will be carried over;
6th digit ---> 25*2+5=55 ---> 5 and 5 will be carried over;
7th digit ---> 24*2+5=53 ---> 3 and 5 will be carried over;
8th digit ---> 23*2+5=51 ---> 1 and 5 will be carried over;
9th digit ---> 22*2+5=49 ---> 9 and 4 will be carried over;
10th digit ---> 21*2+4=46 ---> 6 and 4 will be carried over;
11th digit ---> 20*2+4=44 ---> 4.

Thanks Bunuel for a great and thorough explanation.

Bunuel thanks for the explanation

bunuel

can you please explain this part " 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"

Can someone please clarify what use is the "Sk = Sk–1 + 2(10k–1)" in the stem? is it used to throw you off? I'm not clear as to the reason it was provided nor do I see how it's used to provide the answer, 4.

I understand why the answer is 4, just not why or what that expression is provided for.

For the 10th digit we should add 21 2's --> 21*2=42. Even if carry from the previous operation is 6 (max possible), the sum would be 42+6=48 and the carried over for the next operation will be 4.

Formula \(S_k = S_{k-1} + 2*10^{k-1}\) gives numbers of the sequence: 2, 22, 222, 2222, ...

p is the sum of the first 30 terms of S

p= 2+22+222+2,222+.........+222,222,222,222,222,222,222,222,222,222
= 2(1+11+111+1,111+......+111,111,111,111,111,111,111,111,111,111)...........(A)

1
11
111
1,111
...
111,111,111,111,111,111,111,111,111,111

now in this expression unit digit will be zero (as 1 added 30 times is 30) and the carryover will be 3.

if only ones are to be added the this can be written as (with carrying over anything) and taking the sum of only 11 digits from right to left

..............20 21 22 23 24 25 26 27 28 29 30

now we include carryovers ( carry over from 30 is 3 and (29+3) gives a carryover of 3 and (28+3) gives a carryover of 3 and so on..) and this sum become


.................1 2 3 4 5 6 7 9 0 1 2 0


and now multiplying by 2 that we took out at (A) we get 4 eleventh digit.

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