Carcass wrote:
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4
First recognize that n, n+1 and n+2 are 3 CONSECUTIVE INTEGERS.
Now let's make some observations:
When n = 1, we get: (1)(2)(3), which is NOT divisible by 8
n = 2, we get: (2)(3)(4), which is
DIVISIBLE BY 8n = 3, we get: (3)(4)(5), which is NOT divisible by 8
(4)(5)(6), which is
DIVISIBLE BY 8(5)(6)(7), which is NOT divisible by 8
(6)(7)(8), which is
DIVISIBLE BY 8(7)(8)(9), which is
DIVISIBLE BY 8(8)(9)(10), which is
DIVISIBLE BY 8-----------------------------
(9)(10)(11), which is NOT divisible by 8
(10)(11)(12), which is
DIVISIBLE BY 8(11)(12)(13), which is NOT divisible by 8
(12)(13)(14), which is
DIVISIBLE BY 8(13)(14)(15), which is NOT divisible by 8
(14)(15)(16), which is
DIVISIBLE BY 8(15)(16)(17), which is
DIVISIBLE BY 8(16)(17)(18)which is
DIVISIBLE BY 8-----------------------------
.
.
.
The pattern tells us that 5 out of every 8 products is divisible by 8.
So, 5/8 of the 96 products will be divisible by 8.
This means that the probability is 5/8 that a given product will be divisible by 8.
Answer: D
Cheers,
Brent