Question

Plsss solve it fast......

Show that cube of any positive integer is either of the form 4m, 4m+1, 4m+3 ​

Answer 1

   

Write a few perfect cubes below. First 12 perfect cubes are given below:

1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728...

Divide each term by 4 and write down the remainders thus obtained:

1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0...

This sequence is the repetition of the sequence 1, 0, 3, 0.

Whatever, only 0, 1 and 3 is occurring here.

So the cube of any positive integer is in the form 4m, 4m + 1 and 4m + 3!!!

Hence proved!!!

Plz ask me if you've any doubt on my answer.

Thank you......

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