n is an integer between 1 & 96 chosen at random. What is the probability that n(n+1)(n+2) is divisible by 8.
Apologies I do not have the options for this.
Probability
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Here's the official question with answer choices:
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
-----------------------------
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.
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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
First recognize that n, n+1 and n+2 are 3 CONSECUTIVE INTEGERS.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
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
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n(n+1)(n+2) is a Product of three consecutive Integers which will always be divisible by 8 when n is EVEN (WITH A FEW EXCEPTION THAT WE WILL DISCUSS IN LATER PART OF SOLUTION)akhilsuhag wrote:n is an integer between 1 & 96 chosen at random. 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
CONCEPT : [Because if n is even then n(n+1)(n+2) = (Even)(Odd)(Even) and Two Consecutive Even Integers are always multiple of 8 because one of them will certainly be a multiple of 4 and other will be a multiple of 2 at the least]
From 1 to 96 there are 48 EVEN and 48 ODD Numbers (96/2 = 48)
So possible values of n = 48
EXCEPTION : But n(n+1)(n+2)will also be Divisible by 8 if the number n is Odd but (n+1) itself is a multiple of 8
There are 96/8 = 12 numbers which are multiple of 8 therefore there are 12 more values which (n+1) and hence n can have
Total acceptable values of n = 48+12 = 60
Total Possible values of n = 96
Probability = 56/96 = 5/12
Answer: Option D
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There are situations when n*(n+1)*(n+2),
(1) when two of the numbers are multiples of 2 & 4
n(n+2) is always a multiple of 8 for all even values of n. So there are 48 even numbers between, 0&96,
(2) when one of the three numbers is a multiple of 8
if n=7,15,23,31 ... then (n+1) will be a multiple of 8.
between 0 and 96 there are 12 such numbrs that are one less than a multiple of 8.
so frmo the two cases.. total favourabl outcomes = 48+12=60
so probability of getting a multiple of 8 is 60/96= 5/8
(1) when two of the numbers are multiples of 2 & 4
n(n+2) is always a multiple of 8 for all even values of n. So there are 48 even numbers between, 0&96,
(2) when one of the three numbers is a multiple of 8
if n=7,15,23,31 ... then (n+1) will be a multiple of 8.
between 0 and 96 there are 12 such numbrs that are one less than a multiple of 8.
so frmo the two cases.. total favourabl outcomes = 48+12=60
so probability of getting a multiple of 8 is 60/96= 5/8