Question

use Euclids division lemma to show that the cube of any positive integer is of the form 7m or 7m+1or 7m+6 in short​

Answer 1

Answer:

7m+6

Step-by-step explanation:

Here in this question we have to take b = 7 because if we take b = ( 0 ,1,2,3,4,5,6) then the cube of any of these numbers does not come to be in multiple of 7 so we have the last choice of taking b = 7.

a = 7b+r

r= 0, 1, 2, 3, 4, 5, 6

When r=0

a = 7b

cubing both sides

a3=343b3a3=7(49b3)a3=7m

When r = 1

a = 7b +1

Cubing both sides(CBS)

a3=(7b+1)3a3=343b3+3(49b2)+3(7b)+1a3=7(49b3+21b2+3b)+1a3=7m+1

When r=2

a = 7b +2

CBS

a3=(7b+2)3a3=343b3+3(49b2)2+3(7b)4+8(=7+1)a3=7(49b3+12b2+12b+1)+1a3=7m+1

When r=3

a = 7b + 3

CBS

a3=(7b+3)3a3=343b3+3(49b2)3+3(7b)9+27(=21+6)a3=7(49b3+12b2+12b+3)+6a3=7m+6

Similarly, it can be proved for others.

Note: The value of b taken is generally, the smallest prime number divisor of the coefficient of m. In case, it had been like (9m+1), we would have taken b=3. Here it was (7m+6), so we took b=7.

Taking the smallest prime number divisor reduces values of r and hence the effort.