How do you prove three consecutive numbers are divisible by 6?

For example if you had your three numbers as: 5, 6, 7, one is divisible by 3 and one is divisible by 2, as this is the case with all consecutive three numbers. Therefore as we are multiplying the numbers together, multiplying a multiple of 3 and a multiple of 2 gives us a multiple of 6.

Is the product of three consecutive numbers always divisible by 3?

Therefore, the product of any three consecutive integers is always divisible by 3. There are only two integers between the multiples of 3. So any three consecutive integers will include a multiple of 3. Since the product of the three integers will include a three as a factor, it will be divisible by 3.

What is the product of 3 consecutive even integers?

Explanation: Three consecutive even integers can be represented by x, x+2, x+4. The sum is 3x+6, which is equal to 108. Thus, 3x+6=108.

Which of the following is the greatest divisor of the product of any 3 consecutive even integers?

The largest natural number by which the product of three consecutive even natural numbers is always divisible is 8x2x3 = 48. Hence, the correct option is C.

Why the product of any three consecutive whole numbers is divisible by 6?

Just like the investigation on sum of consecutive numbers we can start by using three consecutive numbers and multiplying them. The answers 6, 24, 60 are all divisible by 6, because each product has an even number and a multiple of 3. Therefore, a number that is always a divisible by 2 and 3 will be divisible by 6.

How do you prove three consecutive numbers are divisible by 3?

If n = 3k + 2, then n + 1 = 3k + 2 + 1 = 3k + 3 = 3(k + 1) which is again divisible by 3. So we can say that one of the numbers among (n, n + 1 and n + 2) is always divisible by 3. Therefore the product of numbers n(n+1)(n+2) is always divisible by 3.

Which one of the following is the greatest number by which the product of the three consecutive even numbers would be exactly divisible?

∴ The number by which the product of three consecutive even numbers would be exactly divisible is 48. In three consecutive numbers there will be two even numbers one a 8 multiple, another number which is 2 multiple and another number which is a 3 multiple. So it will be divisible by 8 x 3 x 2 = 48.

Is the product of three consecutive positive integers divisible by 6?

Prove that the product of three consecutive positive integers is divisible by 6. let the no. 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.

What is the product of three consecutive natural numbers?

Product of three consecutive natural numbers is ( n − 1) n ( n + 1) = n 3 − n. Denote it P. If n is even ( n = 2 k ), then n 3 = 8 k 3 is even and P is even as the difference of two even numbers. Similarly, if n is odd, then n 3 is odd and the difference of two odd numbers is even.

How do you find the product of three numbers?

The three numbers must be integers, and being also consecutive, their units digits must be of the form 0, 1, 2 or 1, 2, 3 or 2, 3, 4 … or 8, 9, 0 or 9, 0, 1. The units digit of their product is equal to the remainder of their product when divided by 10. In our case it must be 4.

Is the number p divisible by 6?

Conclusion: P is divisible by 2 and by 3, therefore is divisible by 6. Out of three consecutive numbers you will always find an even number and a multiple of three. By multiplying, you get a multiple of 6. If n ≡ 1 ( mod 2), n 3 + 3 n ≡ 0 ( mod 2).