Question

class 10th ncert book ex 1.1 ke 2 aur 5 question ka solution

Answer 1

Q.2ANS

let x is greater then 6 by euclid's division lemma we have,
x=6q+r where
0 is greater than equals to r is greater than 6
6q+3 or 6q+1 or 6q+2 or 6q+3 or 6q+4 or 6q+5.
now,6q is an even integer being a multiple of 2
 we know that the sum of even and odd integer is always an odd integer.
therefore, 6q=1 6q+3 and 6q+5 are odd integers

hence any odd positive integer is in the form of 6q+1 or 6q+3 or 6q+5.



Q.5ANS
let x is greater than 3 then by applying euclid's division lemma we get
x=3q+r where 0 is greater than equals to r is greater than 3 

this implies x=3q or 3q+1 or 3q+2
case 1:x=3q
x^3=(3q)3 = 27q^3 =9(3q^3)
=9m where m is some integer

case 2: x=3q=1
this implies x^3=(3q+1)^3=(3q)^3 +3(3q)^2 (1)+3(3q) (1)^2 +1^3
=27q^3 + 27q^2 +9q +1
this implies x^3 =9 (3q^3+3q^2+q)+1
=9m+1 where m is some integer.

case:3 x=3q+3
this implies x^3=(3q+2)^3
=(3q)^3 + 3(3q)^2(2)+3(3q)(2)^3+2^3
=27q^3 + 54q^2+36q+8
=9(3q^3+6q^2+4q)+8
=9m+8 where m is some integer 
hence the cube of any positive integer is in the form of either 9m or 9m +1 or 9m+8.

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