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Re: What is the remainder when 3^243 is divided by 5? [#permalink]
21 Oct 2012, 10:33

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jimhughes477 wrote:

no clue!?!?!

3^1 = 3 3^2 = 9 3^3 = 27 3^4 = 81 3^5 = 243

....

For any power of 3 unit digits would be 1, 3,7,or 9. Also if you notice, after every 4th power of 3, the unit digit would repeat itself. Therefore, in the question 3^243 (or 3^(240+3)) would have unit digit of 7 Hence when divided by 5, it will give remainder =2.

Re: What is the remainder when 3^243 is divided by 5? [#permalink]
22 Oct 2012, 07:14

Expert's post

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jimhughes477 wrote:

What is the remainder when 3^243 is divided by 5?

3^1=3 --> the remainder when we divide 3 by 5 is 3; 3^2=9 --> the remainder when we divide 9 by 5 is 4; 3^3=27 --> the remainder when we divide 27 by 5 is 2; 3^4=81 --> the remainder when we divide 81 by 5 is 1; 3^5=243 --> the remainder when we divide 243 by 5 is 3 AGAIN; ...

As you can see the remainders repeat in blocks of 4: {3, 4, 2, 1}{3, 4, 2, 1}... Since 243=240+3=(multiple of 4)+3, then the remained upon division of 3^243 by 5 will be the third number in the pattern, which is 2.

Re: What is the remainder when 3^243 is divided by 5? [#permalink]
08 Aug 2013, 22:52

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Easier solution:

Identify the numerator value which gives remainder as '1' when divided by '5'

We know that Rem when 81 is divided by 5 is '1'.Also WKT 81 is in powers of '3'

Simplifying

[(3^4)^60 * 3^3]

[(81)^60 * 3^3]

REM of (27/5) =2

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Re: What is the remainder when 3^243 is divided by 5? [#permalink]
11 Aug 2014, 10:45

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