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| 1. |
show that only one of the numbers n , n+2 , n+4 is divisible by 3 |
| Answer» Case-I: When n = 3qIn this case, we haven= 3q, which is divisible by 3Now, n = 3qn+2 = 3q+2n+2 leaves remainder 2 when divided by 3Again, n = 3qn+4 = 3q+4=3(q+1)+1n+4 leaves remainder 1 when divided by 3n+4 is not divisible by 3.Thus, n is divisible by 3 but n+2 and n+4 are not divisible by 3. Case-II: when n = 3q+1In this case, we haven= 3q+1,n leaves remainder 1 when divided by 3.n is divisible by 3Now, n = 3q+1n+2 = (3q+1)+2=3(q+1)n+2 is divisible by 3.Again, n = 3q+1n+4 = 3q+1+4=3q+5=3(q+1)+2n+4 leaves remainder 2 when divided by 3n+4 is not divisible by 3.Thus, n+2 is divisible by 3 but n and n+4 are not divisible by 3. Case-III: When n + 3q+2In this case, we haven= 3q+2n leaves remainder 2 when divided by 3.n is not divisible by 3.Now, n = 3q+2n+2 = 3q+2+2=3(q+1)+1n+2 leaves remainder 1 when divided by 3n+2 is not divisible by 3.Again, n = 3q+2n+4 = 3q+2+4=3(q+2)n+4 is divisible by 3.Thus, n+4 is divisible by 3 but n and n+2 are not divisible 3. | |