IMO: C - 4


unit digit will be added = 30 times = carry on - 6
similarly carry on - 6,6,6,6,5,5,5,5,4,4
so, on the 11th digit we will be adding 20 times 2 + 4 = 4 and 4 carry on...!

how did u get the carried forward # as 6 ?

how did u know that the carried forward # from the 10th digit is 4 and hence 44 ?

am i missing something here ?

can someone elaborate on this method?

is there a quicker approach than this one ?

whats the difficulty level of this sum as per the real GMAT ?

adam15 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?


1
2
4
6
9

the 11 th digit will

30*2------------60----1st dig=0
29*2-----------58+6--2nd dig=4
28*2-------56+6=62---3rd dig=2
27*2----54+6=60------4th dig=0
26*2--52+6=58----5th dig=8
25*2---50+5---6th dig=5
.
.
.
.
.
.
11 th digit is 20*2=40+4=44 is 4

till here u people wud have guessed y i am not scoring good on the D day
this is a typical and a glaring example or way of solving a problem in a layman method
hence conclusion is there is soln for every problem but there is actually no time for solving
experts help with a neat and short method

adam15 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?
1
2
4
6
9

S1 = 2, S2 = 22, S3 = 222,...Sk = Sk-1 + 2(10k-1).
S1 = 2 (1 digit)
S1+S2 = 24(2 digit)
S1 + S2 + S3 = 248(3 digit)
S1 + S2 + S3 + S4 = 2470

....

S1 + S2 + S3 ...S 11 = 2..................xyz

Now what should I do I m totally confused..Please post the OS adam..

The first few terms of the sequence are 2, 22, and 222 and each subsequent term has an additional 2 added on. The 30th term then is a string of 30 2's. If we line up the first 30 terms of the sequence to add them up, we will get rows in the following pattern:

2
22
222
2222
22222
:
:
(30) 2's


To find p, the sum of the first 30 terms of S, we would simply be adding columns of 2's. The key here is to see a pattern in the addition process. Starting with the units digit column, all 30 of the terms have a 2 in that position so the sum of the units column would be 30 x 2 = 60. A zero would be written as the units digit of the sum and a six would be carried over to the tens column.

In the tens column, 29 of the 30 terms would have a 2 because the first term has no tens digit. The sum of the tens digits would be 29 x 2 = 58, to which we must add the 6 for a total of 64. The 4 gets written down as the second digit of p and the 6 is carried over to the hundreds column.

In the hundreds column, 28 of the 30 terms would have a 2, the sum of the hundreds digits would be 28 x 2 = 56, to which we must add the 6 again for a total of 62. The 2 gets written down as the third digit of p and the 6 is carried over to the thousands column.

There are two ways to finish this problem. We can do out the remaining 8 columns and find that the 11th digit (i.e. the 10 billions column) will have a sum of 2(20) + 4 = 44 (where the 4 was carried over from the 10th column). 4 then will be the 11th digit (from the right) of p (and a 4 will be carried over into the 12th column).

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.

Please help urself with the OE

bhumika.k.shah wrote:The first few terms of the sequence are 2, 22, and 222 and each subsequent term has an additional 2 added on. The 30th term then is a string of 30 2's. If we line up the first 30 terms of the sequence to add them up, we will get rows in the following pattern:

2
22
222
2222
22222
:
:
(30) 2's


To find p, the sum of the first 30 terms of S, we would simply be adding columns of 2's. The key here is to see a pattern in the addition process. Starting with the units digit column, all 30 of the terms have a 2 in that position so the sum of the units column would be 30 x 2 = 60. A zero would be written as the units digit of the sum and a six would be carried over to the tens column.

In the tens column, 29 of the 30 terms would have a 2 because the first term has no tens digit. The sum of the tens digits would be 29 x 2 = 58, to which we must add the 6 for a total of 64. The 4 gets written down as the second digit of p and the 6 is carried over to the hundreds column.

In the hundreds column, 28 of the 30 terms would have a 2, the sum of the hundreds digits would be 28 x 2 = 56, to which we must add the 6 again for a total of 62. The 2 gets written down as the third digit of p and the 6 is carried over to the thousands column.

There are two ways to finish this problem. We can do out the remaining 8 columns and find that the 11th digit (i.e. the 10 billions column) will have a sum of 2(20) + 4 = 44 (where the 4 was carried over from the 10th column). 4 then will be the 11th digit (from the right) of p (and a 4 will be carried over into the 12th column).

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.

Please help urself with the OE

bhumika where did you got this original explanation....its really big one...does this type of question gonna come in GMAT...

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?
1
2
4
6
9

when we add 1st 30 numbers, we can see the pattern

22222..........2 (30th term)
..2222..........2 (29th term)
....222..........2 (28th term)
......................
......................
.....................2 (1st term)

____________________

2..2*2..2*3..........2*30


11th digit from right will be 20th digit from left

20th term will be 20*2 =40

4 in 40 will be a carry to 19 th digit sum, and 11th digit from right is now 0

we are left with 0 at 20th position, now carry from sum of 30th digit to 21th digit will decide our answer


there are 10 digits from 30th to 21th

30 will give ---> 30*2= 60 ----(6 will be carry)
29 will give ----29*2 = 58------(5 will be carry) ***numbers will decrease by 2**

remaining 8 numbers are 56, 54, 52, 50, 48, 46, 44, 42

there are one 6, five 5's and four 4's

6*1 +5*5+4*4 = 47

therefore answer should be 4 as 4 will be a carry to 11th digit
.