show that any positive odd integer is of the 6q+1,or6q+3,or 6q+5,where q is some integer please slow paper
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Let take a as any positive integer and b = 6.
Then using Euclid’s algorithm we get a = 6q + 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 6q+0 , 6q+1 , 6q+2,6q+3,6q+4,6q+5
6q+0 6 is divisible by 2 so it is a even number
6q+1 6 is divisible by 2 but 1 is not divisible by 2 so it is a odd number
6q+2 6 is divisible by 2 and 2 is also divisible by 2 so it is a even number
6q+3 6 is divisible by 2 but 3 is not divisible by 2 so it is a odd number
6q+4 6 is divisible by 2 and 4 is also divisible by 2 it is a even number
6q+5 6 is divisible by 2 but 5 is not divisible by 2 so it is a odd number
So odd numbers will in form of 6q + 1, or 6q + 3, or 6q + 5
Then using Euclid’s algorithm we get a = 6q + 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 6q+0 , 6q+1 , 6q+2,6q+3,6q+4,6q+5
6q+0 6 is divisible by 2 so it is a even number
6q+1 6 is divisible by 2 but 1 is not divisible by 2 so it is a odd number
6q+2 6 is divisible by 2 and 2 is also divisible by 2 so it is a even number
6q+3 6 is divisible by 2 but 3 is not divisible by 2 so it is a odd number
6q+4 6 is divisible by 2 and 4 is also divisible by 2 it is a even number
6q+5 6 is divisible by 2 but 5 is not divisible by 2 so it is a odd number
So odd numbers will in form of 6q + 1, or 6q + 3, or 6q + 5
Answered by
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Let the integer b a.
When divided by 6,its form a=6q+r where r is 1,2,3,4,5,0
When r is 0,2,4,Then the members r 6q, 6q+2,6q+4,which r even.
Thus, For being odd, it must b in the form 6q+1,6q+3or 6q+5
When divided by 6,its form a=6q+r where r is 1,2,3,4,5,0
When r is 0,2,4,Then the members r 6q, 6q+2,6q+4,which r even.
Thus, For being odd, it must b in the form 6q+1,6q+3or 6q+5
parveen38:
thanks
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