Question

2. Show that any positive odd integer is of the form ( 6m + 1 ), ( 6m + 3 ) or ( 6m ,+ 5 )

Class 10

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Answer 1

Let us assume that M=q where q is integer

Answer 2

Heya !!!

Let a be a given positive odd integer.

On dividing a by 6 , let m be the quotient and r be the remainder.

Then ,by Euclids division lemma we have

A = 6m + r , where r = 0 , 1 , 2 , 3 , 4 , 5

=> A = 6m [ R = 0 ]

Or,

=> A = 6m + 1 [ R = 1 ]

Or,

=> A = 6m + 2 [ R = 2 ]

Or,

=> A= 6m+ 3 [ R = 3 ]

Or,

=> A = 6m + 4 [ R = 4 ]

Or,

=> A = 6m + 5 [ R = 5 ]

★ A = 6m , A = 6m + 2 , A = 6m + 4 are the even values of A.

When A is odd it is in the form of ( 6m + 1 ) , ( 6m +3 ) and ( 6m +5 ) for some integer m.

★ HOPE IT WILL HELP YOU ★