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n(n+1)(n+2) divisibility by 8

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n(n+1)(n+2) divisibility by 8

by vikz_316 » Wed Nov 26, 2008 10:10 pm
Hello, can anyone help me figure out this problem
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

I am getting the answer as 3/8, but the answer given is 5/8.
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by gmataug08 » Wed Nov 26, 2008 10:29 pm
any 3 consecutive number set having two even numbers would be divisible by 8

eg, 4*5*6 , 12*13*14 , 26*27*28......

96/2 => gives 48 even numbers , hence we will get 48 sets

next , any 3 consecutive number sets having having 2 odd numbers & one 8 or 8 divisible in them would be divisible by 8

eg, 7*8*9 , 15*16*17 .........

96/8 = 12 => that gives 12 sets

total we have 60 sets => 60/96 = 5/8
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by vikz_316 » Wed Nov 26, 2008 10:50 pm
I think i figured it out, but if someone can explain an easier method, i will be extremely grateful.

Here is my method.

maximum value n can have = 96
Values that n,n+1,n+2 can have are
8,16,24,32.......96 = 12 numbers total
7,15,23,31.......95 = 12 numbers total
6,14,22,30.......94 = 12 numbers total
4,12,20,28.......92 = 12 numbers total
2,10,18,26.......90 = 12 numbers total

12*5 = 60 values of n that are possible for which n(n+1)(n+2) will be divisible by 8.
Hence probability = 60/96 = 5/8.
Is there an easier method?
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by vikz_316 » Wed Nov 26, 2008 10:53 pm
thanks a lot my friend, your method definitely is easier than mine. Same idea, but it just didn't strike me. Thanks!
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by gmataug08 » Wed Nov 26, 2008 10:57 pm
glad it helped. :)
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by linfongyu » Thu Mar 19, 2009 8:58 pm
wow. how anyone can solve this - under pressure, in around 2 minutes - is beyond me. to whoever that can, you deserve a cookie.
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Re: n(n+1)(n+2) divisibility by 8

by farooq » Wed Oct 28, 2009 12:30 am
vikz_316 wrote:Hello, can anyone help me figure out this problem
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

I am getting the answer as 3/8, but the answer given is 5/8.
In this questions the probability is results/total outcomes.

n(n+1)(n+2) is a multiple of 8 if,
1. n is a multiple of 8.
2. n+1 is a multiple of 8.
3. n+2 is a multiple of 8.
4. n(n+1) is a multiple of 8.
5. n(n+2) is a multiple of 8.
6. (n+1)(n+2). Ignore this case because we any of them make it a multiple of 8 and we already considered it above.

so from 1...96, we have total 12 numbers that are divisible by 8 or that are multiple of 8.

we have 5 cases, as mentioned above.
therefore total possibilities that n(n+1)(n+2) is

5*12/total experiments.

60/96 = 10/16 =5/8.
Regards,
Farooq Farooqui.
London. UK

It is your Attitude, not your Aptitude, that determines your Altitude.
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by ssmiles08 » Wed Oct 28, 2009 4:31 am
good posts. You can also do this problem by odds and evens.

If n is even, then the product is sure to be divisible by 8 (or three 2's) n is even so it is divisible by one 2. (n+2) is divisible by 4 since it has two 2's.

1/2 of all numbers are even.

If n is odd, (n+1) should be divisible by 8 for the product to be divisible. 96/8 = 12 possible numbers (n+1)

so 12/96 = 1/8 of all numbers are divisible by 8 when n is odd.

so you add both possibilities; when n is even and odd: 1/2 + 1/8 = 5/8
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by Mom4MBA » Fri Jan 22, 2010 1:18 pm
This is long method but easy to understand:

2,3,4
3,4,5
4,5,6
5,6,7
6,7,8
7,8,9 .....................n+1 multiple of 8
8,9,10
9,10,11
10,11,12
11,12,13
12,13,14
13,14,15
14,15,16
15,16,17 ................n+1 multiple of 8
.
.
.
96,97,98

here where n=even; n(n+1)(n+2) is divisible by 8, so we have 96/2 cases

Also see where n+1 is a multiple of 8, the set is divisible by 8, So between 2 - 97 (this is the range for n+1) we have 12 multiples of 8, thus we have 12 more cases

in all 96/2 + 12 = 48+12 = 60 cases are favorable
and total cases are 96

probability = 60/96 = 5/8 Answer
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by asherman » Fri Jan 22, 2010 4:09 pm
Thanks for all the info - it helped me a lot:)
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