If an integer, n, to be chosen at random from 1 to 1000 inclusive, wha

Question

If an integer, n, to be chosen at random from 1 to 1000 inclusive, what is the probability that n(n + 2) will be divisible by 8 ?

A. 1/8
B. 1/4
C. 1/2
D. 2/3
E. 3/4

Bunuel · Accepted Answer

n(n + 2) is NOT divisible by 8 for any odd value of n because the product is odd, and IS divisible by 8 for any even value of n because if n is even, then either n itself or n + 2 will be a multiple of 4, making n(n + 2) divisible by 2 * 4 = 8. Since, from integers from 1 to 1000 inclusive, half are odd and half are even, the probability that n(n + 2) will be divisible by 8 is 1/2.

Answer: C.

$!vakumar.m · Answer

We can see that for odd value of x, n(n+2) is odd integer and for even value of x, n(n+2) is even integer and a multiple of 8.
Therefore we have 50% values divisible by 8.
Answer C

Posted from my mobile device

luisdicampo · Answer

Deconstructing the Question

We want the probability that  n(n+2)  is divisible by  8 , where  n  is chosen uniformly from  1  to  1000 .

The key idea is to use parity.

The numbers  n  and  n+2  always have the same parity.

So if  n  is odd, then both factors are odd, their product is odd, and it cannot be divisible by  8 .

So only even values of  n  can work.

Step-by-step

Let  n  be even.

Write:

n=2k

Then:

n+2=2k+2=2(k+1)

So:

n(n+2)=(2k)\cdot 2(k+1)=4k(k+1)

We want  4k(k+1)  to be divisible by  8 .

That means  k(k+1)  must be divisible by  2 , or in other words,  k(k+1)  must be even.

But  k  and  k+1  are consecutive integers, so one of them is always even.

Therefore  k(k+1)  is always even.

So every even value of  n  works.

From  1  to  1000 , there are  500  even integers out of  1000  total integers.

So the probability is:

\frac{500}{1000}=\frac{1}{2}

Answer: C)  \frac{1}{2}

If an integer, n, to be chosen at random from 1 to 1000 inclusive, wha

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