Question

show that the cube of a positive integer is of the form 6q+r,where q is an integer and r= 0,1,2,3,4,5.




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Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Step-by-step explanation:

Consider a positive integer = 6q + r

where q is an integer and

r = 0, 1, 2, 3, 4, 5.

We know that positive integers are of the form 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4 and 6q + 5.

By taking cube on both sides

For 6q,

(6q)3 = 216q3

= 6(36q)3 + 0

= 6m + 0, (where m is an integer = (36q)3).

For 6q + 1,

(6q + 1)3 = 216q3 + 108q2 + 18q + 1

= 6(36q3 + 18q2 + 3q) + 1

= 6m + 1, (where m is an integer = 36q3 + 18q2 + 3q).

For 6q + 2,

(6q + 2)3 = 216q3 + 216q2 + 72q + 8

= 6(36q3 + 36q2 + 12q + 1) +2

= 6m + 2, (where m is an integer = 36q3 + 36q2 + 12q + 1).

For 6q + 3,

(6q + 3)3 = 216q3 + 324q2 + 162q + 27

= 6(36q3 + 54q2 + 27q + 4) + 3

= 6m + 3, (where m is an integer = 36q3 + 54q2 + 27q + 4).

For 6q + 4,

(6q + 4)3 = 216q3 + 432q2 + 288q + 64

= 6(36q3 + 72q2 + 48q + 10) + 4

= 6m + 4, (where m is an integer = 36q3 + 72q2 + 48q + 10).

For 6q + 5,

(6q + 5)3 = 216q3 + 540q2 + 450q + 125

= 6(36q3 + 90q2 + 75q + 20) + 5

= 6m + 5, (where m is an integer = 36q3 + 90q2 + 75q + 20).

Therefore, the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.

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