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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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New post 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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New post 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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New post 22 Oct 2012, 08:14
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.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html (Please pay attention to the rule #8: Post Answer Choices for PS Questions).
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Re: What is the remainder when 3^243 is divided by 5?  [#permalink]

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

...
where are the answer choices !!!
however, Answer is = +2
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Re: What is the remainder when 3^243 is divided by 5?  [#permalink]

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New post 12 Aug 2014, 00:01
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\(\frac{3^1}{5}\) ..... 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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New post Updated on: 03 Dec 2016, 01:29
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\(\frac{3^{243}}{5}\)

\(3^2\) \(= -1 (mod 5)\)

\(\frac{3^{242}*3}{5} = \frac{(3^2)^{121}*3}{5} = \frac{(-1)^{121}*3}{5} = \frac{-3}{5} = -3 (mod 5) = 2 (mod 5)\)

Remainder is \(2\).

OR

\(3^4 = 1 (mod_5)\)

\(\frac{3^{243}}{5} = \frac{(3^4)^{60}*3^3}{5} = \frac{1^{60}*27}{5}\)

Remainder is \(2\).

Originally posted by vitaliyGMAT on 28 Nov 2016, 08:50.
Last edited by vitaliyGMAT on 03 Dec 2016, 01:29, edited 1 time in total.
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Re: What is the remainder when 3^243 is divided by 5?  [#permalink]

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New post 28 Nov 2016, 10:18
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jimhughes477 wrote:
What is the remainder when 3^243 is divided by 5?


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

\(3^{243}\)

\(= 3 * 3^{242}\)

\(= 3 * 9^{121}\)

\(= 3 * (10 - 1)^{121}\)

On expansion, we will see that except the last term \((-1)^{121} = -1\), all other terms will be divisible by 10 (and hence by 5 too).

Remainder = -3 which is the same as remainder of 2.

For more details on negative remainders, check: https://www.veritasprep.com/blog/2014/0 ... -the-gmat/
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Re: What is the remainder when 3^243 is divided by 5?  [#permalink]

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New post 02 Dec 2016, 06:53
jimhughes477 wrote:
What is the remainder when 3^243 is divided by 5?


To determine the remainder when 3^243 is divided by 5, we need to determine the units digit of 3^243.

Let’s start by evaluating the pattern of the units digits of 3^n for positive integer values of n. That is, let’s look at the pattern of the units digits of powers of 3. When writing out the pattern, notice that we are ONLY concerned with the UNITS digit of each result.

3^1 = 3

3^2 = 9

3^3 = 7

3^4 = 1

3^5 = 3

As we can see from the above, the pattern of the units digit of any power of 3 repeats every 4 exponents. The pattern is 3–9–7–1. In this pattern, all positive exponents that are multiples of 4 will produce a 1 as its units digit. Thus:

3^244 has a units digit of 1, and therefore 3^243 has a units digit of 7. Since 7/5 has a remainder of 2, the remainder when 3^243 is divided by 5 is also 2.
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Re: What is the remainder when 3^243 is divided by 5?  [#permalink]

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New post 03 Jun 2017, 03:49
go for value posting find a patter and answer will be 2
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Re: What is the remainder when 3^243 is divided by 5?  [#permalink]

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New post 03 Jun 2017, 09:23
jimhughes477 wrote:
What is the remainder when 3^243 is divided by 5?


\(\frac{3^4}{5} = Remainder \ 1\)

Now, \(\frac{3^{243}}{5} = \frac{3^{4*60}*3^3}{5}\)

\(3^{4*60}\) will have remainder 1 , when divided by 5
\(3^3\) will have remainder 2 , when divided by 5

So, the Remainder will be 2
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Re: What is the remainder when 3^243 is divided by 5?  [#permalink]

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New post 08 Jun 2017, 23:58
The power cycle of 3 is 4
That is 3^1 is 3 (units digit 3)
3^2 is 9 (units digit 9)
3^3 is 27 (units digit 7)
3^5 is 81 (units digit 1)
3^6 is 243 (units digit 3)
3^7 is 729 (units digit 9)
3^8 is 2187 (units digit 7)
so here we see that the units digit starts with 3 then 9 then 7 and finally 1 and repeats itself.. so any power value f divided by 4.. depending upon the remainder we will be able to tell the units digit of the final no.
===>> we have 3^243 == 243/4 leaves a remainder of 3 ... so we get the final value of 3^243 as blahblah7
blahblah7/5 will get a remainder of 2
Hence 2 is the answer
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Re: What is the remainder when 3^243 is divided by 5?  [#permalink]

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New post 09 Jun 2017, 11:08
Imo 2
Using cyclic properties of three we know the digits repeats in cycle of 4 so 243 /4 we have 3 remainder thus 7 is our units digit which gives 2 remainder on division by 5

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

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New post 19 Jul 2017, 11:07
For any power of 3 unit digits would be 1, 3,7,or 9. Cyclicity is 4.
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.
Please provide the options to verify the solved answer.
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Re: What is the remainder when 3^243 is divided by 5?  [#permalink]

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New post 22 Jul 2017, 08:39
The remainder when we divide 3^exponent by 5 gives a cylicity of 4. Thus 243/4 gives 3 that means 3rd term in the series i.e. 3^3/5 gives remainder as 2. Hence answer should be 2.
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Re: What is the remainder when 3^243 is divided by 5?  [#permalink]

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New post 22 Aug 2017, 00:23
From the cyclicity logic it can be derived that remainder will be 2.
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Re: What is the remainder when 3^243 is divided by 5?  [#permalink]

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New post 22 Aug 2017, 00:37
jimhughes477 wrote:
What is the remainder when 3^243 is divided by 5?

jimhughes477, Please post the options also.

3^243 ~ 3 *3^242/5 ~ 3* 9^121/5 ~ 3* (-1)^121/5 ~ -3/5 ~ 2/5

So, Remainder = 2.
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Re: What is the remainder when 3^243 is divided by 5?  [#permalink]

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New post 04 Sep 2017, 16:25
by using fermat little theorem

a^(p-1)= 1 mod p where a is an integer and p prime
3^(5-1)/5 = 1 mod 5

then 3^243 = 3^(240+3) = 27/5 = 2
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Re: What is the remainder when 3^243 is divided by 5?  [#permalink]

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New post 29 Sep 2018, 11:46
jimhughes477 wrote:
What is the remainder when 3^243 is divided by 5?




\(3^1\) divided by 5 will lead remainder 3
\(3^2\) divided by 5 will lead remainder 4
\(3^3\) divided by 5 will lead remainder 2
\(3^4\) divided by 5 will lead remainder 1
\(3^5\) divided by 5 will lead remainder 3
\(3^6\) divided by 5 will lead remainder 4

The remainder of \(3^n\) divided by 5 has a cyclicity of 4.
so \(3^{243}\) divided by 5 will lead to a remainder of 2
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Re: What is the remainder when 3^243 is divided by 5?  [#permalink]

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Re: What is the remainder when 3^243 is divided by 5?   [#permalink] 16 Oct 2019, 11:49
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