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The remainder of (3^560)/8 is

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The remainder of (3^560)/8 is  [#permalink]

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New post 24 Jun 2019, 03:53
3
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A
B
C
D
E

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Question Stats:

77% (01:21) correct 23% (01:58) wrong based on 48 sessions

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The remainder of (3^560)/8 is
A 1
B 7
C 21
D 27
E 34
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Re: The remainder of (3^560)/8 is  [#permalink]

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New post 24 Jun 2019, 04:16
1
\(3^{560}=(3^2)^{280}=9^{280}\)

\(9\) leaves a remainder of \(1\) when divided by \(8\) and therefore \(9^{280}\) leaves a remainder \(1\) (Remainder of product=Product of Remainders)

Answer is (A)

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New post 24 Jun 2019, 04:19
(3^560)/8 = 3^4(140)/8 =(81^140)/8

81^140/8 —(r)—> 1/8

Since 81/8 has remainder of 1
So (81•81•....140 times)/8 will be
(1•1•1•...140 times) /8

remainder is 1 (A)

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New post 24 Jun 2019, 05:43
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LUCIFER1703 wrote:
The remainder of (3^560)/8 is
A 1
B 7
C 21
D 27
E 34

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Use binomial theorem discussed here:
https://www.veritasprep.com/blog/2011/0 ... ek-in-you/

\(\frac{3^{560}}{8} = \frac{9^{280}}{8} = \frac{(8 + 1)^{280}}{8}\)

Remainder = 1
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New post 24 Jun 2019, 13:55
Mrs Karishma,
The last question(talking about the link you shared) is pretty tricky.

What is the remainder of 2^83 is divided by 9?
2^2*(2^81)= 4*(8^27)/9=4(9-1)^27/9
= 4(9^27+...+(-1)^27)/9.
As we know, the last term of this expression will not be divisible by 9. And we got (-1)^27 which is equal to ( -1).
I got this expression which is m is positive integer.
4*(9m-1)/9= (4*9m-4)/9. And the remainder is 5, not 4. Because of the last term is equal to negative, not positive number.
Am I right?
Thanks in advance!!!

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New post 24 Jun 2019, 14:17
1
\(\frac{3}{8}\) leaves a remainder of 3
\(\frac{3^2}{8}\) = \(\frac{9}{8}\) leaves a remainder of 1
\(\frac{3^3}{8}\) = \(\frac{27}{8}\) leaves a remainder of 3
\(\frac{3^4}{8}\) = \(\frac{81}{8}\) leaves a remainder of 1.

It looks like we have a 3-1 remainder pattern, which means the remainder will repeat twice (multiple of 2), therefore

Remainder of \(\frac{3^{560}}{8}\) = 1 (choice A)
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New post 24 Jun 2019, 14:29
Looks like bebs just beat me to it (great solution!), but there's another option if you aren't feeling confident with the binomial theorem: when numbers are this massive, try finding a pattern. After all, exponents and remainders all relate to number properties (specifically divisibility), and number properties follow patterns.

3^1 = 3 --> 3/8 gives r: 3
3^2 = 9 --> 9/8 gives r: 1
3^3 = 27 --> 27/8 gives r: 3
3^4 = 81 --> 81/8 gives r: 1

At this point, we should be able to see that the patterns of remainders repeats every 2 numbers. In other words, whenever we have an even exponent, the remainder will be 1, and whenever we have an odd exponent, the remainder will be 3. In this problem, our exponent is even, so our remainder will be 1.

This method can also be applied to the problems Karishma provided — for instance, Question 2:

2^1 = 2 --> 2/9 gives r: 2
2^2 = 4 --> 4/9 gives r: 4
2^3 = 8 --> 8/9 gives r: 8
2^4 = 16 --> 16/9 gives r: 7
2^5 = 32 --> 32/9 gives r: 5
2^6 = 64 --> 64/9 gives r: 1
2^7 = 128 --> 128/9 gives r: 2
2^8 = 252 --> 256/9 gives r: 4

This was a harder pattern to find (honestly about as difficult as you can reasonably expect from the GMAT), but we find that the pattern of remainders repeats every 6 numbers. So when the exponent is a multiple of 6 (i.e. we have 2^6n), the remainder will be 1. When the exponent is one higher than a multiple of 6 (i.e. we have 2^(6n+1), the remainder will be 2. When the exponent is two higher than a multiple of 6 (i.e. we have 2^(6n+2), the remainder will be 4. And so on. 86 is 2 higher than 84, the closest multiple of 6. Therefore, the remainder will be 4.

Takeaway: when you're dealing with big numbers and number properties, you can likely find a pattern. Just be sure to stay organized and calculate carefully!
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The remainder of (3^560)/8 is   [#permalink] 24 Jun 2019, 14:29
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