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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.

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]

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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

Rgds, TGC !
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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

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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

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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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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

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

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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

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

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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.

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

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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.