|
Question 1085899: Prove that one and only one out of n,n+4,n+8,n+12 and n+16 is divisible by 5,where n is any pasitive integer
Answer by math_helper(2461)   (Show Source):
You can put this solution on YOUR website!
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
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
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.
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.
|
|
|
| |
