PS

[ketkoag]

If 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)

I got the right answer but i put values and then check and i took more than 3 mins to solve it.
Please lemme know if there is any short and reliable method.
OA: A

[moutar]

If we look at each section of these options modulo 3, they all need to be different.

A. 0 1 2 or 1 2 0 or 2 0 1
B. 0 2 2 or 1 0 0 or 2 1 1
C. 0 0 1 or 1 1 2 or 2 2 0
D. 0 1 1 or 1 2 2 or 2 0 0
E. 0 2 0 or 1 0 1 or 2 1 2

Therefore its A.

[Vemuri]

I will be interested to see any short method to solve this problem. I plugged in numbers greater than 6 & found only A satisying the condition. I took less than 2mins to solve this problem. Basically I was checking for any of the 3 numbers to be multiple of 3 while solving.

[ketkoag]

moutar, could u please elaborate on ur explaination.
vemuri, i did it the same way u did and it took me long coz i double checked each option as it was a bit tricky to use trial and error in the timed conditiond. thanks for ur reply.

[moutar]

You can have 3 types of number.
- a multiple of three (0)
- one more than a multiple of 3 (1)
- two more than a multiple of 3 (2)

The multiplication needs to contain a multiple of 3 (i.e. a zero).

The list contains all the possible options. The only one that is guaranteed to contain a multiple of 3 is A.

[shargaur]

There is rule regarding multiple of 3. If sum of digits are divided by 3 then it is multiple of 3.

A. n (n+1) (n-4) = 3n - 3
B. n (n+2) (n-1) = 3n + 1
C. n (n+3) (n-5) = 3n -2
D. n (n+4) (n-2) = 3n + 2
E. n (n+5) (n-6) = 3n - 1

Except (A) no matter what value of n may be it will never be multiple of 3. it can only be mulitple of 3 if either 3 is subtracted or added to 3n.

Hope it helps

Took less than 20 secs.

[Vemuri]

There is rule regarding multiple of 3. If sum of digits are divided by 3 then it is multiple of 3.

A. n (n+1) (n-4) = 3n - 3
B. n (n+2) (n-1) = 3n + 1
C. n (n+3) (n-5) = 3n -2
D. n (n+4) (n-2) = 3n + 2
E. n (n+5) (n-6) = 3n - 1

Except (A) no matter what value of n may be it will never be multiple of 3. it can only be mulitple of 3 if either 3 is subtracted or added to 3n.

Hope it helps

Took less than 20 secs.

Nice try shargaur. But, don't you think the rule is for adding digits of a number to see if it is a multiple of 3? For ex: 27. We can determine if this number is a multiple of 3 by adding the units & tens digit 2+7 = 9, which is a multiple of 3, hence 27 is a multiple of 3.

When we are dealing with numbers that are multiplying each other (like what is being asked in the question), this rule will not apply.

ex: n(n+2)(n-1). According to what you said, this can be converted to 3n+1 & hence not a multiple of 3. But, when you use a number n=7, this expression is a multiple of 3 (7*9*6), which negates your answer.

Appreciate if you can explain your reasoning.