Please answer and no SPAMMING please. 1) Show that one and only one out of n, (n+2), (n+4) is divisible by 3, where n belongs to N. 2) Show that (n^2-1) is divisible by 8, if n is an odd positive integer.
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1
Answer:
yes
Step-by-step explanation:
Since n, n+1, n+2 are three consecutive integers then there must be one number divisible by 3 at least.
If the remainder at dividing n by 3 is 1, then n+2 must be divisible by 3 and if the remainder at dividing n by 3 is 2, then n+1 must be divisible by 3. Similarly for n+1 and n+2.
Let n be divisible by 3.
3
n+1
=
3
n
+
3
1
Now, n is divisible by 3 but 1 is not. So we get n+1 not divisible by 3. Similarly,n+2 will not be divisible by 3 as well if n is divisible by 3.
3
n+2
=
3
n
+
3
2
In the same way, if n+1 is divisible by 3 then n and n+2 can't be divisible by 3. If n+2 is divisible by 3 then n and n+1 cannot be divisible by 3.
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