Question

use euclid's division Lemma to show that the cube of any positive integer is of the form 9 M, 9M+1 or 9M+8

Answer 1

Hey mate !!

Step by step Explanation :-

• Let the positive integer be a which when divided by 3 gives q as quotient and r as remainder.

by Euclid's division lemma

a=bq+r

a=3q+r

where r=0,1,2

then,

a=3q

or

a=3q+1

or

a=3q+2

now,

• CASE i

a=3q

a³=(3q)³

a³=27q³

a³=9m( where m=3q³)

• CASEii........ using identity

(a+b)³=a³+3a²b+3ab²+b³

a=3q+1

a³=(3q+1)³

a³=27q³+9q²+9q+1

a³=9m(3q³+q²+q)+1

a³=9m+1

• CASE iii

a=3q+2

a³=(3q+2)³

a³=27q³+27q²+36q+8

a³=9m(where m=3q³+3q²+4q)+8

a³=9m+8

Hence , the cube of any +ve no. is of the form 9m , 9m +1 or 9m+8

is shown !!

Answer 2

Step-by-step explanation:

Question : -

→ Use Euclid's Division lemma to show that the Square of any positive integer cannot be of form 5m + 2 or 5m + 3 for some integer m.

$\huge \pink{ \mid{ \underline{ \overline{ \tt Answer: -}} \mid}}$

▶ Step-by-step explanation : -

Let ‘a’ be the any positive integer .

And, b = 5 .

→ Using Euclid's division lemma :-

==> a = bq + r ; 0 ≤ r < b .

==> 0 ≤ r < 5 .

•°• Possible values of r = 0, 1, 2, 3, 4 .

→ Taking r = 0 .

Then, a = bq + r .

==> a = 5q + 0 .

==> a = ( 5q )² .

==> a = 5( 5q² ) .

•°• a = 5m . [ Where m = 5q² ] .

→ Taking r = 1 .

==> a = 5q + 1 .

==> a = ( 5q + 1 )² .

==> a = 25q² + 10q + 1 .

==> a = 5( 5q² + 2q ) + 1 .

•°• a = 5m + 1 . [ Where m = 5q² + 2q ] .

→ Taking r = 2 .

==> a = 5q + 2 .

==> a = ( 5q + 2 )² .

==> a = 25q² + 20q + 4 .

==> a = 5( 5q² + 4q ) + 4 .

•°• a = 5m + 4 . [ Where m = 5q² + 4q ] .

→ Taking r = 3 .

==> a = 5q + 3 .

==> a = ( 5q + 3 )² .

==> a = 25q² + 30q + 9 .

==> a = 25q² + 30q + 5 + 4 .

==> a = 5( 5q² + 6q + 1 ) + 4 .

•°• a = 5m + 4 . [ Where m = 5q² + 6q + 1 ] .

→ Taking r = 4 .

==> a = 5q + 4 .

==> a = ( 5q + 4 )² .

==> a = 25q² + 40q + 16 .

==> a = 25q² + 40q + 15 + 1 .

==> a = 5( 5q² + 8q + 3 ) + 1 .

•°• a = 5m + 1 . [ Where m = 5q² + 8q + 3 ] .

→ Therefore, square of any positive integer in cannot be of the form 5m + 2 or 5m + 3 .

✔✔ Hence, it is proved ✅✅.

$\huge \orange{ \boxed{ \boxed{ \mathscr{THANKS}}}}$