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

\(3^{243}\)

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

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

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

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.

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).

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 !

\(\frac{3^{243}}{5}\)

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

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

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 \(2\).

Cyclicity of 3 is: 3, 9, 7, 1.

243 / 4 = remainder 3.

Therefore units digit of 3^243 is 7.

7 / 5 = remainder 2. Answer is C.

...
where are the answer choices !!!
however, Answer is = +2

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.

go for value posting find a patter and answer will be 2

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

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

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

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

Sent from my ONE E1003 using GMAT Club Forum mobile app

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.

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.

From the cyclicity logic it can be derived that remainder will be 2.

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.

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

\(3^1\) 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