the 11 th digit willadam15 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?
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S1 = 2, S2 = 22, S3 = 222,...Sk = Sk-1 + 2(10k-1).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?
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bhumika where did you got this original explanation....its really big one...does this type of question gonna come in GMAT...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
when we add 1st 30 numbers, we can see the patternS 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?
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