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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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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, 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.

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, 07:14
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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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New post 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]

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New post 09 Aug 2013, 01:49
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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Re: What is the remainder when 3^243 is divided by 5? [#permalink]

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New post 11 Aug 2014, 23:01
\(\frac{3^1}{5}\) ..... 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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What is the remainder when 3^243 is divided by 5? [#permalink]

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New post 28 Nov 2016, 07:50
\(\frac{3^{243}}{5}\)

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

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

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 \(2\).

Last edited by vitaliyGMAT on 03 Dec 2016, 00:29, edited 1 time in total.
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What is the remainder when 3^243 is divided by 5? [#permalink]

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New post 28 Nov 2016, 09:18
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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, 05:53
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] 02 Dec 2016, 05:53
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