APTITUDE MADE EASY

Any power of a number formed by the digit 3 repeated any number of times will give a remainder 3 when divided by 6 

Does it Work? 1.33/6 , Remainder =3 2.999/6 , Remainder = 3 3.2727/6, Remainder = 3 Yes it Does, So Let's Start our Application

1. (9^1+9^2+9^3+9^4+9^5+9^6) /6

We know each power of 9 will give a number of 3. As we are looking at powers of 9 six times. Each one would give 3 as a remainder. So it's (3+3+3+3+3+3) /6 or 6(3) /6 Both of this gives the remainder as 0.

2.(3^1 +33^2+333^3+3333^4........33 terms )/6

We know each of the terms will give a remainder 3. 3^1/6 remainder = 3 33^2/6 1089/6 remainder =3 & so on each gives a remainder 3.

We have 33 terms so it's 33*3 = 99/6 Remainder = 3.

3.(27^1+27^2+27^3+27^4+27^5+27^6.........47 terms )/6

We know each of the terms will give a remainder 3.

27^1/6 remainder = 3 27^2/6 729/6 remainder = 3 & so on each gives a remainder 3. We have 47 terms so it's 47*3 = 141/6

Remainder = 3.

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