show that the product of any three consecutive positive integers is always divisible by 6
Answers
Let the three consecutive positive integers be n, n+1 and n+2.
Whenever a number is divided by 3, the remainder obtained is either 0,1 or 2.
Therefore, n=3p or 3p+1 or 3p+2, where p is some integer.
If n=3p, then n is divisible by 3.
If n=3p+1, then n+2=3p+1+2=3p+3=3(p+1) is divisible by 3.
If n=3p+2, then n+1=3p+2+1=3p+3=3(p+1) is divisible by 3.
So, we can say that one of the numbers among n,n+1 and n+2 is always divisible by 3 that is:
n(n+1)(n+2) is divisible by 3.
Similarly, whenever a number is divided by 2, the remainder obtained is either 0 or 1.
Therefore, n=2q or 2q+1, where q is some integer.
If n=2q, then n and n+2=2q+2=2(q+1) is divisible by 2.
If n=2q+1, then n+1=2q+1+1=2q+2=2(q+1) is divisible by 2.
So, we can say that one of the numbers among n, n+1 and n+2 is always divisible by 2.
Since, n(n+1)(n+2) is divisible by 2 and 3.
Hence, n(n+1)(n+2) is divisible by 6.
Answer:
Step-by-step explanation:
Let the three consecutive positive integers be n, n+1 and n+2.
Whenever a number is divided by 3, the remainder obtained is either 0,1 or 2.
Therefore, n=3p or 3p+1 or 3p+2, where p is some integer.
If n=3p, then n is divisible by 3.
If n=3p+1, then n+2=3p+1+2=3p+3=3(p+1) is divisible by 3.
If n=3p+2, then n+1=3p+2+1=3p+3=3(p+1) is divisible by 3.
So, we can say that one of the numbers among n,n+1 and n+2 is always divisible by 3 that is:
n(n+1)(n+2) is divisible by 3.
Similarly, whenever a number is divided by 2, the remainder obtained is either 0 or 1.
Therefore, n=2q or 2q+1, where q is some integer.
If n=2q, then n and n+2=2q+2=2(q+1) is divisible by 2.
If n=2q+1, then n+1=2q+1+1=2q+2=2(q+1) is divisible by 2.
So, we can say that one of the numbers among n, n+1 and n+2 is always divisible by 2.
Since, n(n+1)(n+2) is divisible by 2 and 3.
Hence, n(n+1)(n+2) is divisible by 6.