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What is the remainder when [m]3^{32}[/m] is divided by 82?

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GreenlightTestPrep
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What is the remainder when [m]3^{32}[/m] is divided by 82? [#permalink] New post   23 Feb 2020, 08:56
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What is the remainder when \(3^{32}\) is divided by \(82\)?

A) 0
B) 1
C) 8
D) 9
E) 81
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B
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Carcass
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Re: What is the remainder when [m]3^{32}[/m] is divided by 82? [#permalink] New post   23 Feb 2020, 11:11
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Good question but I am a witness of the eras: true I am old.

ten years ago a similar question for the gmat or gre was super tough. Now , if you do your home work, it is moderate

See the patter, no need fancy calculations

\(3^2=9\)
\(3^3=27\) (here what count for us is the unit digit = 7
\(3^4=..1\)
\(3^5=..3\)
\(3^6=..9\)

So the patter before to repeat itself is 9-7-1-3-9. We do need the 32nd position7 which is 7

7 divided by 2 (82 we care only of the unit digit) = 1

B is the answer
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Re: What is the remainder when [m]3^{32}[/m] is divided by 82? [#permalink] New post   23 Feb 2020, 11:17
Why 7 is divided by 2? please explain
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Re: What is the remainder when [m]3^{32}[/m] is divided by 82? [#permalink] New post   23 Feb 2020, 11:40
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when you see the pattern the number 3^32 is huge but you need to care only of the unit digit.

the number is xxxxx7 divided by 82

7/2 the remainder is 1

hope now is clear
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Re: What is the remainder when [m]3^{32}[/m] is divided by 82? [#permalink] New post   23 Feb 2020, 12:34
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GreenlightTestPrep wrote:
What is the remainder when \(3^{32}\) is divided by \(82\)?

A) 0
B) 1
C) 8
D) 9
E) 81


If we recognize that 82 is 1 greater than 81 (aka \(3^4\)), then we might see that 82 is a divisor of \(3^{32} - 1\)
Here's why:

\(3^{32} - 1 = (3^{16} + 1)(3^{16} - 1)\)
\(= (3^{16} + 1)(3^8 + 1)(3^8 - 1)\)
\(= (3^{16} + 1)(3^8 + 1)(3^4 - 1)(3^4 + 1)\)
\(= (3^{16} + 1)(3^8 + 1)(3^4 - 1)(82)\)

This tells us that \((3^{32} - 1)\) is a multiple of 82...
...and this means \(3^{32}\) is 1 greater than a multiple of 82, which means we'll get a remainder of 1 when we divide \(3^{32}\) by 82

Answer: B

Cheers,
Brent
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Re: What is the remainder when [m]3^{32}[/m] is divided by 82? [#permalink] New post   Updated on: 22 Sep 2020, 10:57
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Carcass wrote:
Good question but I am a witness of the eras: true I am old.

ten years ago a similar question for the gmat or gre was super tough. Now , if you do your home work, it is moderate

See the patter, no need fancy calculations

\(3^2=9\)
\(3^3=27\) (here what count for us is the unit digit = 7
\(3^4=..1\)
\(3^5=..3\)
\(3^6=..9\)

So the patter before to repeat itself is 9-7-1-3-9. We do need the 32nd position7 which is 7

7 divided by 2 (82 we care only of the unit digit) = 1

B is the answer



Hi Carcass,

As per me,its should be a cycle of 4 not of 3,I am not sure why you are considering units digit as 7 and taking 3rd power of 3.

3 raise to 1 is 3
3 raise to 2 is 9
3 raise to 3 is unit digit 7
3 raise to 4 is unit digit 1

3 raise to 32
32=8x4,i.e,8 cycles of 4

therefore 3 raise to 32 will have unit digit as 1 which when divided by 82 will yield 1

Originally posted by vaishar3 on 20 Sep 2020, 13:21.
Last edited by vaishar3 on 22 Sep 2020, 10:57, edited 1 time in total.
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Re: What is the remainder when [m]3^{32}[/m] is divided by 82? [#permalink] New post   20 Sep 2020, 22:44
pattern starts with 3^0?
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Re: What is the remainder when [m]3^{32}[/m] is divided by 82? [#permalink] New post   21 Sep 2020, 05:26
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no Shubham,

In these questions,always start with power of 1 as it will give the first unit digit in the series of all the powers you are considering.Check my answer above.

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Re: What is the remainder when [m]3^{32}[/m] is divided by 82? [#permalink] New post   05 Jun 2022, 04:26
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