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Why the sum of the three consecutive integers is always divisible by 3?
By the distributive law of multiplication over addition, we can say that the Sum of three consecutive numbers = 3 x (smallest number + 1). Since the right hand side of the equation is divisible by 3, the left hand side must also be! So, the sum of three consecutive numbers is divisible by 3!
Why is the product of three consecutive positive integers is divisible by 6?
Yes, the statement “the product of three consecutive positive integers is divisible by 6” is true. Let us asume the numbers to be (x) , (x + 1) ,(x + 2). A number which is divided by 3, will be having the remainder 0 or 1 or 2. so, we can say that one of the numbers n, n + 1 and n + 2 is always divisible by 3.
Which among the following is a product of three consecutive numbers whose mean is a perfect cube?
Then [4, 5, 6] is a sequence of 3 consecutive integers that each produce a perfect cube since f(4) = f(5) = f(6) = 8 which is a perfect cube.
Is the product of three consecutive integers always a multiple of 3?
As out of three consecutive integers one of them is always a perfect multiple of 3 and hence the product of the consecutive integers is also a multiple of three . Let us say 3 consecutive numbers are x,x+1,x+2 . When it’s multiplied by 3, then the result is divisible by 3 always.
How many consecutive integers are divisible by 3?
Thus, of any three consecutive integers exactly one of them will be divisible by 3, but that is enough to guarantee that their product will be divisible by 3. In this regard, what is the product of three consecutive integers? So the product of three consecutive integers will be a multiple of 3.
How do you prove a number is divisible by 3?
Case 1: x is positive integer. Then the number formed becomes x (x+1) (x+2) and the sum of the digits of this number is always the multiples of 3, i.e, it is 3*k where k is positive integer, hence the number is divisible by 3. Notice that the formed number will be at least a three-digit number (for 1<=x<=7) or more.
How do you find the sum of three consecutive integers?
Let the three consecutive integers be x, x+1, and x+2 such that: Case 1: x is positive integer. Then the number formed becomes x(x+1)(x+2) and the sum of the digits of this number is always the multiples of 3, i.e, it is 3*k where k is positive integer,…