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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
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:
Re: 1 + 2 + 2^2 + 2^3 + ......... + 2^40 = N
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15 Oct 2023, 09:15
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