Question

a) Can a divisor of 6 have a remainder of 5?....
in
f 12​

Answer 1

Answer:

Yes,..

Solution: 6k+5≡?(mod12)

Now, see the interesting pattern

For k=1,6k+5=11≡11(mod12)

For k=2,6k+5=17≡5(mod12)

For k=3,6k+5=23≡11(mod12)

⋮

So, remainder alternates between5 and 11. To be specific, remainder is 11 if k is odd and 5 if k is even. It is obvious for even as we may write k=2m, m∈N for even, ⟹

For k=2m,6×2m+5=12m+5≡5(mod12)

as 12|12m and remainder is 5 .