341c≡0 is obvious. Now look at the (n-1) numbers c+a, c+2a,...,c+(n-1)a. Suppose any two have the same residue then their difference should be divisible by n. Prove this cannot happen.
Prove::::::::
If gcd(a,n)=1,then the integers c,c+a,c+2a,c+3a,...........................,c+(n-1)a form a complete set of reidues modulo n, for any c.
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6 Answers
341
39anirban, just state what u mean by the question in a simpler way
i am not a dumb guy tto ask u to explain your question
put your question in mathematical terms
i would be even greatful if u tell me whats your approach
1@prophet sir
Perhaps this will seem to u a dumb q but I cannot understand why c≡0 is obvious.It is not mentioned that c is divisible by n
1Let S = { c , c + a , c + 2 a . . . . . . }
Let us assume , contrary to the the hypothesis , that S does not form a complete set of residues
modulo n . Then there exist 2 distinct positive integers " d " and " c " with b < d < n ( without loss
of generality ) such that ,
c + a + d ≡ c + a + b ( mod n )
It follows that , d ≡ b ( mod n )
Which obviously is a contradiction since , b < d < n implies that ------ d - b < n .
QED .