Question

iv) Each question carties 4 Marks.
14. A) Show that the cube of any positive integer is in the form of either 5m or 5m + 1 or
5m + 2 or 5m + 3 or 5m + 4.​

Answer 1

Answer:

It need to be proved. The proving is given below:

Step-by-step explanation:

Let x be any positive integer. Then, it is of the form 5q or, 5q+1 or, 5q+2 or,5q+3 or,5q+4. So, we have the following cases:

CASE 1 When x = 5q : In this case we have

             $x^{3}$ = $(5q)^{3}$ = $125q^{3}$ = 5$(25q^{3})$ = 5m, where m=$25q^{3}$

CASE 2 When x = 5q + 1 : In this case we have

             $x^{3}$ = $(5q+1)^{3}$

             $x^{3}$ = $125q^3 + 75q^2 + 15q + 1$

             $x^{3}$ = $5q(25q^2 + 15q + 3) + 1$

             $x^{3}$ = $5m + 1, where m=q(25q^2 + 15q + 3)$

CASE 3 When x = 5q + 2 : In this case we have

             $x^{3}$ = $(5q + 2)^3$

             $x^{3}$ = $125q^3 + 75q^2 + 15q + 8$

             $x^{3}$ = $5q(25q^2 + 15q + 3) + 8$

             $x^{3}$ = 5m + 2, where m$=q(5q^2 + 15q + 3) + 6$

CASE 4 When x = 5q + 3 : In this case we have

             $x^{3}$ = $(5q + 3)^3$

             $x^{3}$ = $125q^3 + 75q^2 + 15q + 27$

             $x^{3}$ = $5q(25q^2 + 15q + 3) + 27$

             $x^{3}$ = $5m + 3, where m=q(25q^2 + 15q + 3) + 24$

CASE 5 When x = 5q + 4 : In this case we have

             $x^{3}$ = $(5q + 4)^3$

             $x^{3}$ = $125q^3 + 75q^2 + 15q + 64$

             $x^{3}$ = 5q$(25q^2 + 75q + 3 + 64)$

             $x^{3}$ = $5m+4, where m=q(25q^2 + 15q + 3) + 60$

Hence x is either of the 5q or, 5q+1 or, 5q+2 or,5q+3 or,5q+4.

                          THANK YOU

Answer 2

Answer:

Just see weather this helps you