Question

show that any positive odd integer is of the form 6 m + 1 or 6 m + 3 or 6 m + 5 Where M is some integer​

Answer 1

Answer:

Step-by-step explanation:

Let take a as any positive integer and b = 6. a > b

Then using Euclid’s algorithm, we get a = 6m + r here r is remainder and value of q is more than or equal to 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < b and the value of b is 6

So total possible forms will be 6m + 0, 6, + 1 , 6m + 2,6m + 3,6m + 4,6m + 5.

6m + 0 → 6 is divisible by 2 so it is an even number.

6m + 1 → 6 is divisible by 2 but 1 is not divisible by 2 so it is an odd number.

6m + 2 → 6 is divisible by 2 and 2 is also divisible by 2 so it is an even number.

6m + 3 → 6 is divisible by 2 but 3 is not divisible by 2 so it is an odd number.

6m + 4 → 6 is divisible by 2 and 4 is also divisible by 2 it is an even number.

6m + 5 → 6 is divisible by 2 but 5 is not divisible by 2 so it is an odd number.

So odd numbers will in form of 6m + 1, or 6m + 3, or 6m + 5.

Answer 2

HEY MATE HERE IS UR ANSWER

Let n be a given positive odd integer.

On dividing n by 6 , let m be the Quotient and r be the remainder.

Then, by Euclid division lemma, we have

Dividend = Divisor × Quotient + Remainder

》n = 6m + r. where r = 0 , 1 ,2 , 3 ,4 ,5

》n= 6m + 0 = 6m. [ r =0]

》n = 6m + 1 = 6m+1 [ r = 1 ]

》n = 6m +2 = 6m +2 [ r = 2 ]

》n = 6m +3 = 6m+3 [ r = 3 ]

》n = 6m +4 = 6m+4 [ r = 4 ]

》n = 6m+5 = 6m+5 [ r = 5 ]

N = 6m , (6m +2) , (6m+4) is even value of n.

Therefore,

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