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1 + 2 + 2^2 + 2^3 + ......... + 2^40 = N
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27 Aug 2021, 09:27
27 Aug 2021, 09:27
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\(1 + 2 + 2^2 + 2^3 + ......... + 2^{40} = N\)
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
Remainder when \(N\) is divided by 9 |
4 |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
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1 + 2 + 2^2 + 2^3 + ......... + 2^40 = N
[#permalink]
29 Aug 2021, 07:38
29 Aug 2021, 07:38
2
*This problem is too tough and too lengthy for GRE*
\(1 + 2 + 2^2 + 2^3 + ......... + 2^{40} = N\)
This is a Geometric Series with First term, a = 1 and common ratio ,r as 2 and number of terms n as 41. [ \(2^0\) to \(2^{40}\) ]
[ Watch this video to learn about Geometric Series ]
Quantity A:
Using Sum formula for Geometric Series
\(S_{n} = a*\frac{((๐^๐ โ ๐)}{(๐ซโ๐))}\)
N = \(1*\frac{((2^{41} โ ๐)}{(2โ๐))}\) = \(2^{41}\) - 1
Now, we need to find remainder of \(2^{41}\) - 1 by 9
We can do this by finding out the cycle of remainder of power of 2 by 9 [ Watch this video to learn this trick ]
\(2^1\) by 9 remainder is 2
\(2^2\) by 9 remainder is 4
\(2^3\) by 9 remainder is 8
\(2^4\) by 9 remainder is 7
\(2^5\) by 9 remainder is 5
\(2^6\) by 9 remainder is 1 [ Since we got 1 so it will repeat from next one ]
\(2^7\) by 9 remainder is 2
So, we need to divide the power of 2 by 6 (the cycle) and check the remainder
\(41/6\) remainder is 5
=> \(2^{41}\) will give the same remainder by 9 as \(2^5\)
=> Remainder of \(2^{41}\) by 9 is 5
Remainder of -1 by 9 will be 8 [-1 +9 = 8]
So, total remainder of \(2^{41}\) - 1 by 9 = 5 + 8 =13. This cannot be more than 9 so need to divide again by 9 to get final remainder as
\(13/9\) = 4 = Quantity B
Hence, Answer will be C
Hope it helps!
Video on Sequences
\(1 + 2 + 2^2 + 2^3 + ......... + 2^{40} = N\)
This is a Geometric Series with First term, a = 1 and common ratio ,r as 2 and number of terms n as 41. [ \(2^0\) to \(2^{40}\) ]
[ Watch this video to learn about Geometric Series ]
Quantity A:
Using Sum formula for Geometric Series
\(S_{n} = a*\frac{((๐^๐ โ ๐)}{(๐ซโ๐))}\)
N = \(1*\frac{((2^{41} โ ๐)}{(2โ๐))}\) = \(2^{41}\) - 1
Now, we need to find remainder of \(2^{41}\) - 1 by 9
We can do this by finding out the cycle of remainder of power of 2 by 9 [ Watch this video to learn this trick ]
\(2^1\) by 9 remainder is 2
\(2^2\) by 9 remainder is 4
\(2^3\) by 9 remainder is 8
\(2^4\) by 9 remainder is 7
\(2^5\) by 9 remainder is 5
\(2^6\) by 9 remainder is 1 [ Since we got 1 so it will repeat from next one ]
\(2^7\) by 9 remainder is 2
So, we need to divide the power of 2 by 6 (the cycle) and check the remainder
\(41/6\) remainder is 5
=> \(2^{41}\) will give the same remainder by 9 as \(2^5\)
=> Remainder of \(2^{41}\) by 9 is 5
Remainder of -1 by 9 will be 8 [-1 +9 = 8]
So, total remainder of \(2^{41}\) - 1 by 9 = 5 + 8 =13. This cannot be more than 9 so need to divide again by 9 to get final remainder as
\(13/9\) = 4 = Quantity B
Hence, Answer will be C
Hope it helps!
Video on Sequences
1 + 2 + 2^2 + 2^3 + ......... + 2^40 = N
[#permalink]
02 Sep 2021, 18:59
02 Sep 2021, 18:59
1
Definitely a tough question. A full understanding of how remainders work is needed.
I agree with BrushMyQuant; I wouldn't expect to find this on the GRE. However, if you do find the answer, you can come away from this knowing you have a very solid grasp of the topic.
Essentially, all you need to understand is that you can add remainders if the dividends are divided by the same divisor.
For example, let's use 24 and 35 as dividends, and 9 as the divisor, and see how the remainders behave:
\(\frac{24}{9}\) gives a remainder of 6
\(\frac{35}{9}\) gives a remainder of 8
\(24+35 = 59\)
\(\frac{59}{9}\) gives a remainder of 5
Now let's add the remainders from before and see what happens:
\(6 + 8 = 14\)
\(\frac{14}{9}\) gives a remainder of 5.
So if we find the remainder of each individual term in the sum, sum up these remainders, and then divide that result by the divisor, it would be the same as finding the remainder if we sum the entire sequence and then divide that by the divisor.
So looking at the original question we have:
\(1 + 2 + 2^2 + 2^3 + .... + 2^{40}\)
If we divide this sum by 9, and split up each individual term, we get:
\(\frac{1}{9} + \)\(\frac{2}{9} + \)\(\frac{2^{2}}{9} + \)\(\frac{2^{3}}{9} + \)....\(\frac{2^{40}}{9}\)
Let's simplify the first few terms to get an idea of what's going on:
\(\frac{1}{9} + \)\(\frac{2}{9} + \)\(\frac{4}{9} + \)\(\frac{8}{9} + \)....\(\frac{2^{40}}{9}\)
The patter, as BrushMyQuant showed above, repeats every 6th term:
\(\frac{1}{9}\) has remainder 1
\(\frac{2}{9}\) has remainder 2
\(\frac{4}{9}\) has remainder 4
\(\frac{8}{9}\) has remainder 8
\(\frac{16}{9}\) has remainder 7
\(\frac{32}{9}\) has remainder 5
\(\frac{64}{9}\) has remainder 1
.
.
.
The number of terms in our sequence is 41, since we have to count the first term, which is \(2^0\) (40-0+1 = 41)
This means that this pattern will repeat 6 times, and then go another 5 terms before ending. (6*6 + 5)
Using the property I detailed before beginning the question, let's sum the first 6 remainders.
\(1+2+4+8+7+5 = 27\)
Since the pattern repeats itself six times, we'll get 27 six times. The sum of the remaining 5 remainders will be:
\(1+2+4+8+7 = 22\)
So we have:
\(27 + 27 + 27 + 27 + 27 + 27 + 22\)
Like in the property above, let's divide these by 9 and find their remainders:
\(\frac{27}{9} + \)\(\frac{27}{9} + \)\(\frac{27}{9} + \)\(\frac{27}{9} + \)\(\frac{27}{9} + \)\(\frac{27}{9} + \)\(\frac{22}{9}\)
The remainders will be:
\(0 + 0 + 0 + 0 + 0 + 0 + 0 + 4\)
\(4\)
Therefore the answer is C
___________
In short, the property detailed above should be understood, as it might appear on the GRE, but a question like this will not be.
I agree with BrushMyQuant; I wouldn't expect to find this on the GRE. However, if you do find the answer, you can come away from this knowing you have a very solid grasp of the topic.
Essentially, all you need to understand is that you can add remainders if the dividends are divided by the same divisor.
For example, let's use 24 and 35 as dividends, and 9 as the divisor, and see how the remainders behave:
\(\frac{24}{9}\) gives a remainder of 6
\(\frac{35}{9}\) gives a remainder of 8
\(24+35 = 59\)
\(\frac{59}{9}\) gives a remainder of 5
Now let's add the remainders from before and see what happens:
\(6 + 8 = 14\)
\(\frac{14}{9}\) gives a remainder of 5.
So if we find the remainder of each individual term in the sum, sum up these remainders, and then divide that result by the divisor, it would be the same as finding the remainder if we sum the entire sequence and then divide that by the divisor.
So looking at the original question we have:
\(1 + 2 + 2^2 + 2^3 + .... + 2^{40}\)
If we divide this sum by 9, and split up each individual term, we get:
\(\frac{1}{9} + \)\(\frac{2}{9} + \)\(\frac{2^{2}}{9} + \)\(\frac{2^{3}}{9} + \)....\(\frac{2^{40}}{9}\)
Let's simplify the first few terms to get an idea of what's going on:
\(\frac{1}{9} + \)\(\frac{2}{9} + \)\(\frac{4}{9} + \)\(\frac{8}{9} + \)....\(\frac{2^{40}}{9}\)
The patter, as BrushMyQuant showed above, repeats every 6th term:
\(\frac{1}{9}\) has remainder 1
\(\frac{2}{9}\) has remainder 2
\(\frac{4}{9}\) has remainder 4
\(\frac{8}{9}\) has remainder 8
\(\frac{16}{9}\) has remainder 7
\(\frac{32}{9}\) has remainder 5
\(\frac{64}{9}\) has remainder 1
.
.
.
The number of terms in our sequence is 41, since we have to count the first term, which is \(2^0\) (40-0+1 = 41)
This means that this pattern will repeat 6 times, and then go another 5 terms before ending. (6*6 + 5)
Using the property I detailed before beginning the question, let's sum the first 6 remainders.
\(1+2+4+8+7+5 = 27\)
Since the pattern repeats itself six times, we'll get 27 six times. The sum of the remaining 5 remainders will be:
\(1+2+4+8+7 = 22\)
So we have:
\(27 + 27 + 27 + 27 + 27 + 27 + 22\)
Like in the property above, let's divide these by 9 and find their remainders:
\(\frac{27}{9} + \)\(\frac{27}{9} + \)\(\frac{27}{9} + \)\(\frac{27}{9} + \)\(\frac{27}{9} + \)\(\frac{27}{9} + \)\(\frac{22}{9}\)
The remainders will be:
\(0 + 0 + 0 + 0 + 0 + 0 + 0 + 4\)
\(4\)
Therefore the answer is C
___________
In short, the property detailed above should be understood, as it might appear on the GRE, but a question like this will not be.
