Problem

What is the remainder when 2⁵⁵  + 1 is divided by 33.?

The simple way to solve this problem is to reduce the number

conveniently smaller to workout.When 1000 is divided by 33, the remainder will be 10 ,If it is 1024, the remainder will be 10+24=34, and the final remainder will be 1

2⁵⁰ = (2¹⁰)⁵ = (1024)⁵ = (y+1)⁵ , where y is perfectly divisible by 33.On expanding it into a power series,all the terms will contain y except the last term which will be equal to 1

2⁵⁵ = (2¹⁰)⁵ x 2⁵ = (y+1)⁵ x 32 .The ultimate remainder will be 32 times the remainder in (y+l)⁵.i.e., 32 x 1 = 32.

Alternatively it can be worked out by considering 2⁵⁵ = (2¹¹)⁵ = (2048)⁵

 The remainder for 2000 is 20 and for 2048 it is 68 and final remainder is 2

2⁵⁵ = (y+2)⁵, where y is perfectly divisible by 33. The polynomial technique predicts  its last term (2⁵) as remainder .It is 33 when + 1 is also taken into account.

Problem

What is the remainder when (918273)³ when divided by 7 ?

One may know that all cube numbers when divided by 7, the remainder will always be 1 or 6 or 0.

When the base number is divisible by 7, the remainder will be 0, It the base number is in the form (1+7n) or(2+7n) or (4+7n), the remainder will be 1 ,If it is in the form (3+7n) or (5+7n) or (6+7n), the remainder will be 6.

918273/7 =131181(7) +6, so the remainder is 6.

It shows that the remainder of any base number is equal to the remainder of cube of the number when divided by 7.