prove that the product of every three consecutive integers is always divisible by 3 as well as 2.
Asked by Amyra B | 17th Apr, 2013, 06:47: PM
Expert Answer:
Lets take three consecutive integers as n , n+1 and n+2. Now, there are 2 possibilities -
1. Two numbers are odd and one is even.
2. Or two numbers are even and one is odd.
So the even number (irrespective of the fact that there would be 1 or 2 even numbers) is always divisible by two.
And one of the odd numbers is divisible by three (remember you are taking three consecutive numbers and every third integer is a number series is divisible by 3). Therefore the product is divisible by 2 as well as 3.
Lets take three consecutive integers as n , n+1 and n+2. Now, there are 2 possibilities -
1. Two numbers are odd and one is even.
2. Or two numbers are even and one is odd.
So the even number (irrespective of the fact that there would be 1 or 2 even numbers) is always divisible by two.
And one of the odd numbers is divisible by three (remember you are taking three consecutive numbers and every third integer is a number series is divisible by 3). Therefore the product is divisible by 2 as well as 3.
Answered by | 17th Apr, 2013, 08:53: PM
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