Question

Use Euclid’s division lemma to show that the cube of any positive integer is of the form 7m, 7m+1, 7m+6.

Answer 1

Hint: Euclid’s division lemma is (a = b $ \times $q + r). Consider b as 7 and then r gets the values less than 7 (0, 1, 2…6). Now take the cube on both sides and then take common terms that relate to the ones in question.

Complete step-by-step answer:
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
$
  {a^3} = 343{b^3} \\
  {a^3} = 7(49{b^3}) \\
  {a^3} = 7m \\
 $
When r = 1
a = 7b +1
Cubing both sides(CBS)
$
  {a^3} = {(7b + 1)^3} \\
  {a^3} = 343{b^3} + 3(49{b^2}) + 3(7b) + 1 \\
  {a^3} = 7(49{b^3} + 21{b^2} + 3b) + 1 \\
  {a^3} = 7m + 1 \\
$
When r=2
a = 7b +2
CBS
$
  {a^3} = {(7b + 2)^3} \\
  {a^3} = 343{b^3} + 3(49{b^2})2 + 3(7b)4 + 8( = 7 + 1) \\
  {a^3} = 7(49{b^3} + 12{b^2} + 12b + 1) + 1 \\
  {a^3} = 7m + 1 \\
$
When r=3
a = 7b + 3
CBS
$
  {a^3} = {(7b + 3)^3} \\
  {a^3} = 343{b^3} + 3(49{b^2})3 + 3(7b)9 + 27( = 21 + 6) \\
  {a^3} = 7(49{b^3} + 12{b^2} + 12b + 3) + 6 \\
  {a^3} = 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.