**Question:**

Find the remainder when 3^27 is divided by 5?

Okk..thanx esa qus first time dekha..so confuse thi...thanx 4 explanation

Rt and 2,

Manisha 3 to the power 27 h.

Sol:- {(3^4)^6}×3^3

={81^6}×27

=1×27 (81^6 ka unit digit= 1)

so 27÷5 ka remainder 2 hoga

3^27 =(9^13)×3

wen divided by 5 we get

( 4^13) × 3

4 = (5-1)

by binomial theorem we get,,

(-1)^13 + 3= 2

3

3^27= ((3^4)^6) * (3^3) = (81^6) * 27 then unit digit of (81^6) is 1 so on multiplying with 27,

unit digit in the result will be 7. now, 7 when divided by 5 gives 2 as remainder

3^1=3 >remainder =3

3^2=9 >remainder =4

3^3=27>remainder =2

3^4=81>remainder =1

3^5=243>remainder =3

From here, we notice the remainder repeats in a fashion of degree 4.

Therefore 3^27= 3^24*3^3=(3^4)^6*3^3

After 3^24 , the remainder will repeat again.

So, for remaining term(3^3), the remainder is 2.

Hence, the remainder of 3^27 is also 2.

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