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

Question

S is the infinite sequence  S_1 = 2, \ S_2 = 22, \ S_3 = 222, \ ..., \ S_k = 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) 6
(E) 9

Bunuel · Accepted Answer

S is the infinite sequence  S_1 = 2, \ S_2 = 22, \ S_3 = 222, \ ..., \ S_k = 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) 6
(E) 9

--------------------------------------------------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.

mbaiseasy · Answer

1st digit: 2*30 = 60, carry over 6
2nd digit: 2*29 = 58 + 6 = 64, carry over 6
3rd: 2*28 = 56 + 6 = 62, carry over 6
4th: 2*27 = 54 + 6 = 60, carry over 6
5th: 2*26 = 52 + 6 = 58, carry over 5
6th: 2*25 = 50 + 5 = 55, carry over 5
7th: 2*24 = 48 + 5 = 53, carry over 5
8th: 2*23 = 46 + 5 = 51, carry over 5
9th: 2*22= 44 + 5 = 49, carry over 4
10th: 2*21 = 42 + 4 = 46, carry over 4
11th: 2*20 = 40 + 4 = 44

Answer: 4

seekmba · Answer

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.

Bunuel · Answer

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.

seekmba · Answer

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

SonyGmat · Answer

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?

Bunuel · Answer

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.

enigma123 · Answer

Thanks Bunuel for a great and thorough explanation.

subhajeet · Answer

Bunuel thanks for the explanation

ratinarace · Answer

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"

xchakax · Answer

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.

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

Bunuel · Answer

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.

Bunuel · Answer

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

stoned · Answer

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.

akshatv45 · Answer

Using proper sequence functions:

S for 30 elements = S1 + S2 + S3 .... + S30

S1 = 2, S2 = 2 + 2(10), S3 = 2 + 2(10) + 2(10^2) and so on.

So Essentially, the sum of 30 elements is 2*30 + 2*10*29 + 2*(10^2)*28 + 2*(10^3)*27 + ........ + 2*(10^29)*1

Expanding the above sum, you will see a pattern, it starts with 60 and reduces by 2 after that.

60
 580
 5600
 54000
 520000
 5000000
 48000000
 460000000
 4400000000
 42000000000
 400000000000
--------------------
When you sum all this, clearly the 11th digit is 4

Kinshook · Answer

S is the infinite sequence  S_1 = 2, \ S_2 = 22, \ S_3 = 222, \ ..., \ S_k = 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?

p_1 = S_1 = 2; 11th digit = 0
p_2 = S_1 + S_2 = 2 + 22 = 24; 11th digit = 0
p_3 = 246; 11th digit = 0
p_4 = 2468; 11th digit = 0
p_5 = 24690; 11th digit = 0
p_6 = 246912; 11th digit = 0
p_7 = 2469134; 11th digit = 0
p_8 = 24691356; 11th digit = 0
p_9 = 246913578; 11th digit = 0
p_10 = 2469135800; 11th digit = 0
p_11 = 24691358022; 11th digit = 2
p_12 = 246913580244; 11th digit = 4
p_13 = 2469135802466; 11th digit = 6
p_14 = 24691358024688; 11th digit = 9
p_15 = 246913580426910; 11th digit = 1
p_16 = 2469135804269132; 11th digit = 3
p_17 = 24691358042691354; 11th digit = 5
p_18 = 246913580426913576; 11th digit = 8
p_19 = 2469135804269135798; 11th digit = 0
p_20 = 24691358042691358020; 11th digit = 4
p_21 = 246913580426913580242; 11th digit = 2
p_22 = 2469135804269135802464; 11th digit = 6
p_23 = 24691358042691358024686; 11th digit = 9
p_24 = 246913580426913580246908; 11th digit = 1
p_25 = 2469135804269135802469230; 11th digit = 3
p_26 = 24691358042691358024692452; 11th digit = 5
p_27 = 246913580426913580246924674; 11th digit = 8
p_28 = 2469135804269135802469246906; 11th digit = 0
p_29 = 24691358042691358024692469128; 11th digit = 2
p_30 = 246913580426913580246924691350;11th digit = 4

IMO C

Kinshook · Answer

S is the infinite sequence  S_1 = 2, \ S_2 = 22, \ S_3 = 222, \ ..., \ S_k = 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?

1st digit: 2*30 = 60, carry over 6 
 2nd digit: 2*29 = 58 + 6 = 64, carry over 6 
 3rd: 2*28 = 56 + 6 = 62, carry over 6 
 4th: 2*27 = 54 + 6 = 60, carry over 6 
 5th: 2*26 = 52 + 6 = 58, carry over 5 
 6th: 2*25 = 50 + 5 = 55, carry over 5 
 7th: 2*24 = 48 + 5 = 53, carry over 5 
 8th: 2*23 = 46 + 5 = 51, carry over 5 
 9th: 2*22= 44 + 5 = 49, carry over 4 
 10th: 2*21 = 42 + 4 = 46, carry over 4 
 11th: 2*20 = 40 + 4 = 44

IMO C

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

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards