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

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What is the remainder when 3^243 is divided by 5? [#permalink]

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21 Oct 2012, 11:25
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What is the remainder when 3^243 is divided by 5?
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Re: What is the remainder when 3^243 is divided by 5? [#permalink]

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21 Oct 2012, 11:33
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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.

Hope it helps.
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Re: What is the remainder when 3^243 is divided by 5? [#permalink]

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22 Oct 2012, 08:14
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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.

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Re: What is the remainder when 3^243 is divided by 5? [#permalink]

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08 Aug 2013, 23:52
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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]

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09 Aug 2013, 02:49
jimhughes477 wrote:
What is the remainder when 3^243 is divided by 5?

...
where are the answer choices !!!
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Re: What is the remainder when 3^243 is divided by 5? [#permalink]

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11 Aug 2014, 11:45
Hello from the GMAT Club BumpBot!

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Re: What is the remainder when 3^243 is divided by 5? [#permalink]

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12 Aug 2014, 00:01
$$\frac{3^1}{5}$$ ..... Remainder = 3

$$\frac{3^2}{5}$$ ..... Remainder = 4

$$\frac{3^3}{5}$$ ..... Remainder = 2

$$\frac{3^4}{5}$$ ..... Remainder = 1

$$\frac{3^5}{5}$$ ..... Remainder = 3 & so on

So, the cyclicity of remainder is 3,4,2,1.........

$$\frac{3^{243}}{5}$$ .... Remainder would be same as $$\frac{3^3}{5}$$ ..... Remainder = 2
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Re: What is the remainder when 3^243 is divided by 5? [#permalink]

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31 Aug 2015, 02:58
Hello from the GMAT Club BumpBot!

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Re: What is the remainder when 3^243 is divided by 5?   [#permalink] 31 Aug 2015, 02:58
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