FDP Question Bank #14

[Kevin]

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 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.



My question is with regards to the second method explained for solving the problem. The explanation states that the 11th column has 20 2's for a total of 40. I follow this. With regards to the carry over from the 10th column, it is stated that it can't be more then 8. How do we know this and if it is 8 then it will lead us to a different answer then 44, correct?

Also, why can't the 9th column inherit 8 from the 9th column? Thanks

[RonPurewal]

My question is with regards to the second method explained for solving the problem. The explanation states that the 11th column has 20 2's for a total of 40. I follow this. With regards to the carry over from the 10th column, it is stated that it can't be more then 8. How do we know this and if it is 8 then it will lead us to a different answer then 44, correct?

here's one way to get the '8' thing:
in all thirty numbers, the contents of the 10 right-hand columns is 2,222,222,222 or less. (to be precise, it's exactly 2,222,222,222 for the last twenty-one of the numbers, and less than that for the first nine.) since all of these are less than 3,000,000,000, the sum of those ten columns is less than 30 x 3,000,000,000 = 90,000,000,000. therefore, the carry digit is less than nine; in other words, 8 or less.

it's a very crude approximation, but it's good enough.

Also, why can't the 9th column inherit 8 from the 9th column? Thanks

the explanation above takes into account ALL of the carries, because it involves the actual sum of all the numbers in ALL the places to the right of the 11th place. therefore, no further consideration is necessary.

[ColinT819]

Just out of curiosity how many people actually identified how to solve this and then did it under 2 minutes? This to me seems like a skip question!

[RonPurewal]

Just out of curiosity how many people actually identified how to solve this and then did it under 2 minutes?

"two minutes" is an AVERAGE, so this is a non-issue.
if you have decent time management, you should take more than the average time nearly as often as you take less.

[RonPurewal]

...and this thread is now locked, since it's in the wrong folder (there's a separate folder for questions from our own sources).