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GRE Prep Club Team Member
Joined: 20 Feb 2017
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S is the infinite sequence S1 = 2, S2 = 22, S3 = 222,...Sk = Sk–1 + 2
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04 Apr 2021, 22:43
04 Apr 2021, 22:43
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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?
A) 1
B) 2
C) 4
D) 6
E) 9
A) 1
B) 2
C) 4
D) 6
E) 9
Re: S is the infinite sequence S1 = 2, S2 = 22, S3 = 222,...Sk = Sk–1 + 2
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10 Apr 2021, 02:27
10 Apr 2021, 02:27
solution plz
S is the infinite sequence S1 = 2, S2 = 22, S3 = 222,...Sk = Sk–1 + 2
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Updated on: 12 Apr 2021, 13:25
Updated on: 12 Apr 2021, 13:25
3
I kinda did it in a long way, not sure this is the right approach to the problem.
First, I'm going to explain the approach for a small series, let's do that for only 5 terms of the given series.
S1 = 2, S2 = 22, S3 = 222, S4 = 2222, S5 = 22222, S6 = 222222 .... S30 = 222222.....2222
2
22
222
2222
22222
Look closely, for the unit digit [right most] there will be 5 of 2's, So 5 * 2 = 10 and 0 is the unit digit, and 1 is carry forward.
then, for the tenth digit, it will be 2 * 4 = 8 + 1 (from the carry forward of unit digit's calculation)
So, the whole point is to sum the value of multiples of 2's and a single carry forward.
Now, let's consider the whole scenario here,
at most right position you will have 30 of 2's So, 30 * 2 = 60, the position value is 0, and carry forward is 6
then come to one position right, now you have 29 of 2's and a previous carry forward of 6. So, 29 * 2 = 58 + 6 = 64, position value will be 4 and carry forward is now 6.
This may seem tedious calculation, but when you do that on paper it's just simple multiplication and addition. By following the same process 19 times you will reach the 11th position, that will be our answer.
First, I'm going to explain the approach for a small series, let's do that for only 5 terms of the given series.
S1 = 2, S2 = 22, S3 = 222, S4 = 2222, S5 = 22222, S6 = 222222 .... S30 = 222222.....2222
2
22
222
2222
22222
Look closely, for the unit digit [right most] there will be 5 of 2's, So 5 * 2 = 10 and 0 is the unit digit, and 1 is carry forward.
then, for the tenth digit, it will be 2 * 4 = 8 + 1 (from the carry forward of unit digit's calculation)
So, the whole point is to sum the value of multiples of 2's and a single carry forward.
Now, let's consider the whole scenario here,
at most right position you will have 30 of 2's So, 30 * 2 = 60, the position value is 0, and carry forward is 6
then come to one position right, now you have 29 of 2's and a previous carry forward of 6. So, 29 * 2 = 58 + 6 = 64, position value will be 4 and carry forward is now 6.
This may seem tedious calculation, but when you do that on paper it's just simple multiplication and addition. By following the same process 19 times you will reach the 11th position, that will be our answer.
