2: Consecutive numbers and divisibility by 3 and 6:On a number line every nth number is divisible by n (n, 2n, 3n….)

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E.g.: Every 3rd number is divisible by 3 (3, 6, 9…)

Every 4th number is divisible by 4 (4, 8, 12..)

As every third number is divisible by 3, any number n on the number line:

-> Is either, divisible by 3 (it is of the form 3k)

-> Or, leaves a remainder 1 when divided by 3 (it is of the form 3k + 1) - in this case n+2 will be divisible by 3

-> Or, leaves a remainder 2 when divided by 3 (it is of the form 3k+2) - in this case n+1 will be divisible by 3

Hence, the product of three consecutive numbers n(n+1)(n+2) will always be divisible by 3.

As discussed in the previous post, the product of three consecutive numbers is always divisible by 2. So, the product of three consecutive numbers will always be divisible by 2*3 = 6 as well.

Takeaway: a product of three consecutive numbers is always divisible by 3 and 6.We can also say that the sum of the digits of the product is divisible by 3 (a number is only divisible by 3 if the sum of its digits is divisible by 3)

Now, let’s consider the product of three numbers of the form:

(n +/- 3k)(n +1 +/- 3l)(n + 2 +/- 3m), where k, l and m are integers.

As you can see, a multiple of three is added to the individual numbers in the product of consecutive integers n(n+1)(n+2).

For example: 4 * 8 * 6

It is of the form (4) * (5 + 3) * (6) - > 3 is added to the middle number in the product of consecutive numbers 4*5*6

(4) * (5 + 3) * (6) = 4 * 5 * 6 + 4 * 3 * 6

This is clearly divisible by 3

(n +/- 3k)(n +1 +/- 3l)(n + 2 +/- 3m) = n(n+1)(n+2) + terms divisible by 3

As the product of three consecutive numbers is always divisible by 3, the above product is divisible by 3.

Takeaway: the product of three numbers of the form (n +/- 3k)(n +1 +/- 3l)(n + 2 +/- 3m) is always divisible by 3.You can use this property for solving

OG 13 PS Question 87If n is an integer greater than 6, which of the following must be divisible by 3 ?

(A) n(n+1)(n-4)

(B) n(n+2)(n-1)

(C) n(n+3)(n-5)

(D) n(n+4)(n-2)

(E) n(n+5)(n-6)

Option (A) - n(n+1)(n-4) = n(n+1) (n+2 - 6) (as n>6, this product will be positive)

As you can see this product is of the form n(n+1)(n+2 - 3*2)

It is divisible by 3.

Answer: Option (A)

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