Question

prove that every cube integer is in the form of 3k or 3k+1​

Answer 1

We'd divide 12 into 50 to get a quotient of 4 with a remainder of 2. So we could make 4 cartons of eggs with 2 left over for breakfast. It would not make sense to have a remainder that was 12 or more because then we'd just make more cartons until we had less than 12 left over.

This can be expressed by the equation: 50 = 4 * 12 + 2,where 4 is the quotient and 2 is the remainder.

So since dividing an integer n by 3 must leave a remainder less than 3, the remainder must be 0 or 1 or 2.

If the remainder is 0, this can be written as the equation:

n = 3k where k is the quotient, and 0 is the remainder. (You can write this as n = 3k + 0 to see the remainder explicitly).

If the remainder is 1, this can be written as the equation:

n = 3k + 1 where k is the quotient, and 1 is the remainder.

And if the remainder is 2, this can be written as the equation:

n = 3k + 2 where k is the quotient, and 2 is the remainder.

This takes care of every case, thus every integer is of the form 3k, or 3k + 1, or 3k + 2.

Answer 2

Step-by-step explanation:

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