tarek99 wrote:

If n is an integer greater than 6, which of the following must be divisible by 3?

a) n(n+2)(n+3)

b) n(n+5)(n-3)

c) n(n+2)(n+5)

d) n(n+1)(n-4)

e) n(n+1)(n-2)

I chose A as my answer, but it's wrong. How come??? whenever we have 3 consecutive numbers, obviously one of them will always be a multiple of 3. I really don't understand why it is wrong this time!

I chose D too..

here is the strategy

...(n-3)(n-2) (n-1) ..... n (n+1) (n+2) .... (n+3)(n+4)(n+5)..

--> multiple of any three consecutive positive numbers defnitiely divsiable by 3.

n (n+1) (n+2) --- > must divisible by 3

(n-6)(n-5)(n-4)...(n-3)(n-2) (n-1) ..... n (n+1) (n+2) .... (n+3)(n+4)(n+5)..

for any one number if substract or add 3(or multiples of 3) ..then it must divisible by 3.

n (n+1) (n+2 -3)= n (n+1) (n-1)

n (n+1) (n+2 -3*2)= n (n+1) (n-4)

n (n+1) (n+2 +3)= n (n+1) (n+5)

n (n+1) (n+2 +3*2)= n (n+1) (n+8)

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By above strategy (substraction or addition .. to any numbers) I try to see if any answer choice will lead to three consecutive numbers..

a) n(n+2)(n+3) -- out not consecutive numbers...

b) n(n+5)(n-3) --> n (n+2) n --out not consecutive numbers...

subtract 3 from (n+5) add 3 to (n-3)

c) n(n+2)(n+5) -->n(n+2)(n+2) --out not consecutive numbers

subtract 3 from (n+5)

d) n(n+1)(n-4) -->n(n+1)(n-1) -- consecutive numbers add 3 to (n-4)

e) n(n+1)(n-2) --> n(n+1)(n+1) out not consecutive numbers

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