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MGMAT Question

by Strongt » Sat Jan 14, 2012 3:23 pm
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

Answer is c

is there a simple/effecient way to solve this question in less than 10 minutes
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by GMATGuruNY » Sat Jan 14, 2012 9:49 pm
Strongt 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

Answer is c

is there a simple/effecient way to solve this question in less than 10 minutes
Calculate just enough to see the pattern.

The UNITS digit of p will be determined by the sum of 30 2's: 30*2 = 60.
The TENS digit of p will be determined by whatever is carried over from the units place and the sum of 29 2's: 29*2 = 58.
The HUNDREDS digit of p will be determined by whatever is carried over from the tens place and the sum of 28 2's: 28*2 = 56.

The results are DECREASING CONSECUTIVE EVEN INTEGERS.

So the sums that will determine the digits of p, from left to right, are:
40-42-44-46-48-50-52-54-56-58-60.

Now work from right to left to see what gets carried over from one sum to the next.
In the list below, each integer in red represents a digits of p; the other value gets carried over to the next sum.
60 is in the units place, so the units digit is 0.
Carry the 6.
58+6 = 64.
Carry the 6.
56+6 = 62.
Carry the 6.
54+6 = 60.
Carry the 6.
52+6 = 58.
Carry the 5.
50+5 = 55.
Carry the 5.
48+5 = 53.
Carry the 5.
46+5 = 51.
Carry the 5.
44+5 = 49.
Carry the 4.
42+4 = 46.
Carry the 4.
40+4 = 44.
The 11th digit from the right = 4.

The correct answer is C.
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by Strongt » Sat Jan 14, 2012 10:22 pm
Thanks GMATGuruNY

Appreciate you taking the time to answer this but i've been trying to figure it out for 20 minutes!

I got the 1st part where you identified the consecutive even integers.

But I'm confused regarding the numbers carried over to the next sum!

Thank you
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by ArunangsuSahu » Sat Jan 14, 2012 10:32 pm
without carry sum of the 2's for the 11th digit=20*2=40
4 gets carried from the 10th positio.

40+4=44

so resulting digit=4
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by GMATGuruNY » Sun Jan 15, 2012 7:30 am
Strongt wrote:Thanks GMATGuruNY

Appreciate you taking the time to answer this but i've been trying to figure it out for 20 minutes!

I got the 1st part where you identified the consecutive even integers.

But I'm confused regarding the numbers carried over to the next sum!

Thank you
Please see the following illustration:
Image

Units place:
The sum of the 30 2's is 60.
60 units = 6 TENS and 0 units.
Thus, the units digit is 0, and the 6 TENS are carried over to the TENS place.

Tens place:
When we add the 6 tens carried over from the units place to the sum of the 29 2's, we get 6+58 = 64.
64 tens = 6 HUNDREDS and 4 TENS.
Thus, the tens digit is 4, and the 6 HUNDREDS are carried over to the HUNDREDS place.

Hundreds place:
When we add the 6 hundreds carried over from the tens place to the sum of the 28 2's, we get 6+56 = 62.
62 hundreds = 6 THOUSANDS + 2 HUNDREDS.
Thus, the hundreds digit is 2, and the 6 THOUSANDS are carried over to the THOUSANDS place.

And so on.
The rest of the calculations are illustrated here:
Image

The 11th digit from the right is 4.
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