Question 1085899

Without loss of generality, assume n mod 5 = 0   (i.e. assume it is 'n' that 5 divides evenly.  It will be clear why this doesn't change anything):
Then…
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n mod 5 = 0
(n+4) mod 5 = 4
(n+8) mod 5 = 3
(n+12) mod 5 = 2
(n+16) mod 5 = 1
——————
if we continued along larger values of n…
(n+20) mod 5 = 0  
(n+24) mod 5 = 4
(n+28) mod 5 = 3
(n+32) mod 5 = 2
(n+36) mod 5 = 1
——————
and continuing toward smaller values of n…
(n-4) mod 5 = 1
(n-8) mod 5 = 2
(n-12) mod 5 = 3
(n-16) mod 5 = 4
(n-20) mod 5 = 0
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Now,  if it were any other value 5 divided evenly (say n+12 mod 5 = 0) then just let m = n+12  and now the problem hasn't changed: m mod 5 = 0 and m replaces n in our table.
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Another, simpler way to look at is is  that  given  n, n+4, n+8, n+12, and n+16,   each will give one of the remainders  0,1,2,3,4  when divided by 5.