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

Question

A. 0
B. 1
C. 2
D. 3
E. 4

Vips0000 · Accepted Answer

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.

Bunuel · Answer

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).

TGC · Answer

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

REM of (27/5) =2

Rgds,
TGC !

Asifpirlo · Answer

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

PareshGmat · Answer

\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

vitaliyGMAT · Answer

\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 .

KarishmaB · Answer

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.

JeffTargetTestPrep · Answer

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.

srinidhi mishra · Answer

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

Abhishek009 · Answer

\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

harsha3699 · Answer

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

arvind910619 · 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

FB2017 · Answer

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.

FB2017 · 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.

FB2017 · Answer

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

shashankism · Answer

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.

cbh · Answer

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

Princ · Answer

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

Basshead · Answer

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.

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?

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