nourelhoudask wrote:
Bunuel wrote:
AbdurRakib wrote:
If k is a positive integer, what is the remainder when (k + 2)(k^3 – k) is divided by 6 ?
A. 0
B. 1
C. 2
D. 3
E. 4
There is a rule saying that: The product of n consecutive integers is always divisible by n!.
\((k + 2)(k^3 – k) = (k + 2)k(k^2 – 1) = (k + 2)k(k – 1)(k+1)=(k – 1)k(k+1)(k + 2)\).
As we can see we have the product of 4 consecutive integers: (k – 1), k, (k+1), and (k + 2). Thus, the product will always be divisible by 4! = 24, which means that it will be divisible by 6 too.
Answer: A.
Of course we can simply plug-in numbers: since PS question cannot have more than 1 correct answer, then for any k the answer must be the same. For example, if k = 1, then the product becomes 0. 0 divided by 6 gives the remainder of 0 (0 is a multiple of every integer).
Hello Bunnel,
Thank you for your answer, can you please explain why the product of 4 consecutive integers will always be divisible by 4?
Thank you
Hi nourelhoudask,
The concept that you are asking about is based on a couple of Number Property rules - and you can prove it rather easily if you want to by TESTing VALUES.
For a positive integer to be 'evenly divisible by 4' means that the integer is a 'multiple of 4'; in other words, you have to be able to find a '4' when you prime-factor that number (meaning that you have to find at least two 2s) in its prime-factorization.
For example, 12 is evenly divisible by 4 because when you prime-factor 12, you get (2)(2)(3).
In that same way, 15 is NOT evenly divisible by 4 because when you prime-factor 15, you get (3)(5) ---> we don't have the two 2s that we need.
When you multiply four consecutive integers together, you will ALWAYS have a multiple of 4 in that product. For example:
(1)(2)(3)(4) = (6)(4) = 24 --> a multiple of 4
(2)(3)(4)(5) = (30)(4) = 120 --> a multiple of 4
(3)(4)(5)(6) = (90)(4) = 360 --> a multiple of 4
(4)(5)(6)(7) = (210)(4) = 840 --> a multiple of4
That pattern will continue; you will just be dealing with larger and larger multiples of 4 as you go:
(5)(6)(7)(8) = (210)(8) = (420)(4) = 1680 --> a multiple of 4
(6)(7)(8)(9)
(7)(8)(9)(10)
(8)(9)(10)(11)
(9)(10)(11)(12)
Etc.
This shows that every group of four consecutive positive intergers will always include a multiple of 4, so the product of that group will - by definition - be a multiple of 4.
GMAT assassins aren't born, they're made,
Rich
Contact Rich at: Rich.C@empowergmat.com