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S is the infinite sequence S1=2, S2=22, S3=222 [#permalink]
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18 Jul 2010, 15:47
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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 Anyone knows what is the easiest and fastest method of solving this ?
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Re: Infinite seq of 2 [#permalink]
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mn2010 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?
A) 1 B) 2 C) 4 D) 6 E) 9
Anyone knows what is the easiest and fastest method of solving this ? 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. Just curios: is it GMAT question?
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Re: Infinite seq of 2 [#permalink]
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21 Jul 2010, 13:58
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.



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Re: Infinite seq of 2 [#permalink]
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21 Jul 2010, 14:21
seekmba wrote: 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. 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.
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Re: Infinite seq of 2 [#permalink]
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22 Jul 2010, 07:43
Thanks Bunuel. Makes sense. I forgot about the evenly spaced set and hence mixed it up.



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Re: Infinite seq of 2 [#permalink]
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13 Jan 2012, 16:55
Bunuel wrote: mn2010 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?
A) 1 B) 2 C) 4 D) 6 E) 9
Anyone knows what is the easiest and fastest method of solving this ? 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. Just curios: is it GMAT question? 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?



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Re: Infinite seq of 2 [#permalink]
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13 Jan 2012, 17:18
SonyGmat wrote: Bunuel wrote: mn2010 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?
A) 1 B) 2 C) 4 D) 6 E) 9
Anyone knows what is the easiest and fastest method of solving this ? 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. Just curios: is it GMAT question? 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? 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.
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Re: S is the infinite sequence S1=2, S2=22, S3=222 [#permalink]
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13 Jan 2012, 23:12
Thanks Bunuel for a great and thorough explanation.
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Re: S is the infinite sequence S1=2, S2=22, S3=222 [#permalink]
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14 Jan 2012, 03:17
Bunuel thanks for the explanation



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Re: S is the infinite sequence S1=2, S2=22, S3=222 [#permalink]
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21 Dec 2012, 08:06
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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
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Re: S is the infinite sequence S1=2, S2=22, S3=222 [#permalink]
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21 Dec 2012, 08:35
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"



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Re: S is the infinite sequence S1=2, S2=22, S3=222 [#permalink]
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21 Dec 2012, 17:28
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.



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Re: S is the infinite sequence S1=2, S2=22, S3=222 [#permalink]
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23 Dec 2012, 07:15



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Re: S is the infinite sequence S1=2, S2=22, S3=222 [#permalink]
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23 Dec 2012, 07:18



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Re: S is the infinite sequence S1=2, S2=22, S3=222 [#permalink]
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24 Dec 2012, 22:25
mn2010 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?
A) 1 B) 2 C) 4 D) 6 E) 9
Anyone knows what is the easiest and fastest method of solving this ? The answer is C i.e. 4. The 11th place from left has twenty 2's. Sum of which is 40, note there is a carry of 4 from the 10th digit from left and thus the total sum is 44. Thus the 11th digit from left is 4. I have used the long conventional way to figure this out. It took a good 3 mins. Can anyone share a way where you can find trends in the data. Thanks



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Re: S is the infinite sequence S1 = 2, S2 = 22, S3 = 222,...Sk = [#permalink]
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08 Jun 2013, 01:56
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here is a good manual solution and it works... sistheinfinitesequences12s222s3222s33128.html#p227101ps_dahiya wrote: C
Sum of unit digits of first 30 terms = 60 Sum of tens digits of first 30 terms = 58 Sum of thousands digits of first 30 terms = 56 and so on..
p1 = 0 p2 = (6+58) = 4 p3 = (6+56) = 2 p4 = (6+54) = 0 p5 = (6+52) = 8 p6 = (5+50) = 5 p7 = (5+48) = 3 p8 = (5+46) = 1 p9 = (5+44) = 9 p10= (4+42) = 6 p11= (4+40) = 4: ANSWER Carry over is added to next sum.
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Re: S is the infinite sequence S1 = 2, S2 = 22, S3 = 222,...Sk = [#permalink]
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08 Jun 2013, 02:07
S = 2,22,222,2222 ........ 30 terms Sum = 2+22+222+2222+.........+30 terms Let unit digit of sum = 2* 30 =60 (zero is unit digit) Let Tens digit = 2* 29 +6(carry) = 64 (4 is tens digit)
To find 11th digit from right end we do 2*20 + x(carry from 10th digit) = 40 +x
To find x lets find 10th digit= 2*21 +y(carry from 9th digit) For Eg: unit digit got carry = 6, then y should be less than/equal to 6 therefore 10th digit be 42+6 or 42 +5 or (42 + some positive number less than/equal to 6). So for 11th digit carry be 4 Then 11th digit be (40 + 4 =44) = 4
OA = C(4)



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