If an integer n is 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
A.1/4
B.3/8
C.1/2
D.5/8
E.3/4
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If an integer n to be chosen randomly between 1 and 96 inclusive, what is the probability that n(n+1)(n+2) is divisible by 8 ?
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4
OA: D
Any time you have a divisibility question, think of it in terms of the factors you need. Since we need (n)(n+1)(n+2) to divide by 8, we need three factors of 2 SOMEWHERE. We can do this in a few ways:Abhijit K wrote:If an integer n is 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
... and that's a much easier question to answern is an integer from 1 to 96, inclusive. What's the probability that n is either even or one less than a multiple of 8?
Brent@GMATPrepNow wrote:If an integer n to be chosen randomly between 1 and 96 inclusive, what is the probability that n(n+1)(n+2) is divisible by 8 ?
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4
OA: D
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 8
n = 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, [spoiler]5/8[/spoiler] 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
Divisibility is a common topic. See here for some more examples:Abhijit K wrote:If an integer n is 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