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Probability Problem

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Probability Problem

by gmat009 » Tue Oct 07, 2008 12:54 pm
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

Oa is D
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Re: Probability Problem

by Morgoth » Tue Oct 07, 2008 1:22 pm
gmat009 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

Oa is D
n(n + 1)(n + 2) is a product of 3 consecutive integers.

if n is even, n(n + 1)(n + 2), will be divisible by 8.

even integers from 1-96 inclusive = (96-2)/ 2 = 94/2 = 47, 47+1 = 48

also if n is odd and 1 less than multiple 8, n(n + 1)(n + 2) will be divisible by 8, because this will have at least 1 multiple of 8.

multiples of 8 from 1-96 = (96-8)/8 = 11, 11+1 = 12


total probability = 48+12 = 60/96 = 10/16 = 5/8.

Hope this helps.
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Re: Probability Problem

by gmat009 » Tue Oct 07, 2008 2:38 pm
Morgoth wrote:
gmat009 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

Oa is D
n(n + 1)(n + 2) is a product of 3 consecutive integers.

if n is even, n(n + 1)(n + 2), will be divisible by 8.

even integers from 1-96 inclusive = (96-2)/ 2 = 94/2 = 47, 47+1 = 48

also if n is odd and 1 less than multiple 8, n(n + 1)(n + 2) will be divisible by 8, because this will have at least 1 multiple of 8.

multiples of 8 from 1-96 = (96-8)/8 = 11, 11+1 = 12


total probability = 48+12 = 60/96 = 10/16 = 5/8.

Hope this helps.
Thanks for explanation
I am not clear with just 1 thing.
You calculated " even integers from 1-96 divisible by 2=48"
and "multiples of 8 from 1-96 =12"
Total 48+12 but they will not be having some common numbers which are even and divisible by 8 and we are counting them twice.
Don't we need to subtract those numbers.
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Re: Probability Problem

by Morgoth » Tue Oct 07, 2008 4:55 pm
gmat009 wrote: Thanks for explanation
I am not clear with just 1 thing.
You calculated " even integers from 1-96 divisible by 2=48"
and "multiples of 8 from 1-96 =12"
Total 48+12 but they will not be having some common numbers which are even and divisible by 8 and we are counting them twice.
Don't we need to subtract those numbers.
No. what I did was find the odd integers which lead to one 8 in n(n+1)(n+2)

lets start with odd numbers that will lead to one 8 in n(n+1)(n+2)

n=7, 7*8*9
n=15, 15*16*17
n=23, 23*24*25
so on and so forth

We basically need to find the how many odd numbers for n will satisfy the above condition.

All numbers 1 less than the multiple of 8 will satisfy the above condition. Therefore, if we find the multiples of 8, we have found the exact number of odd integers.

Hope this helps.
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Re: Probability Problem

by bullshark » Wed Jul 08, 2009 5:14 pm
Morgoth wrote:
gmat009 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

Oa is D
n(n + 1)(n + 2) is a product of 3 consecutive integers.

if n is even, n(n + 1)(n + 2), will be divisible by 8.

even integers from 1-96 inclusive = (96-2)/ 2 = 94/2 = 47, 47+1 = 48

also if n is odd and 1 less than multiple 8, n(n + 1)(n + 2) will be divisible by 8, because this will have at least 1 multiple of 8.

multiples of 8 from 1-96 = (96-8)/8 = 11, 11+1 = 12


total probability = 48+12 = 60/96 = 10/16 = 5/8.

Hope this helps.
can you explain why you do (96-2)/2? then you add one to the result?
why did you not simply do 96/2 = 48 and 96/8 = 12 ?

Thanks
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Re: Probability Problem

by Morgoth » Thu Jul 09, 2009 4:45 am
bullshark wrote: can you explain why you do (96-2)/2? then you add one to the result?
why did you not simply do 96/2 = 48 and 96/8 = 12 ?

Thanks
I did that only to be safe. Here I could have simply done the division and it still would have worked because the starting number was 1. Had it not been 1 or some other whole number starting from 30s or 40s, the simply division method would have missed 2 important values and would have lead to the wrong answer.
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Still confused..

by uw490 » Thu Jul 09, 2009 7:51 am
Perhaps someone could help clear this up still...

I understand where the 12 comes from in the calculations (multiples of 8), but still am confused where we get 48. Unless I've misread the problem, and the subsequent explanations 48 represents all the other even numbers? But obviously, numbers such as 20, 22, 36, etc are not divisible by 8. This is where I lose the explanation of the problem.

Any clarification on this portion of the problem would be great.

thanks

J
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n(n+1)(n+2)

by bbaah » Thu Jul 09, 2009 8:37 am
We are interested in numbers (n) such that n(n+1)(n+2) is divisible by 8.

if n=22, then 22(22+1)(22+2)= 22(23)(24): This product is divisible by 8 because it has a factor of 2 in 22= 11*2 and a factor of 4 in 24 = 4*6.

A similar explanation can be given for n = 24, 26, etc.

Hope this helps.
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by cameronwu » Thu Jul 09, 2009 8:56 am
48 comes from the fact that every single even number works in this equation, and here's why.

Take n=2. It'll end up being 2 * (3) * (4)... now if you factor this out, you'll see that you get 2^1 * 3 * 2^2 .... This leads to 2^3 = 8*3.

Take n=4.... 4*5*6.... same thing as 2^2 * 5 * (2*3)...another 2^3 in there.

If you take the pattern out further, you'll see you'll be to extract a 2^3 factor for EVERY SINGLE EVEN NUMBER.

There are 48 even numbers from 1-96 inclusive. n*(n+2) with even numbers will always give you a multiple of 2^3 in there.

The 12 comes from the (n+1) term. It's every number that's a multiple of 8, minus 1. Like 7, 15, 23, etc... all the way through 95. If you counted all those terms up, then it would come to 12 terms.

12 odd terms + 48 even terms = 60 total N's that work.
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Makes sense...

by uw490 » Thu Jul 09, 2009 11:10 am
Ok, gotcha.

Makes sense, but... (I'm probably making this way more difficult than I need to)

How do we know to look for the "number that's a multiple of 8, minus 1. Like, 7, 15, 23, etc..." Still not clear why we know to look for this portion of the solution.

bear with me

thanks!

J
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Re: Makes sense...

by cameronwu » Thu Jul 09, 2009 12:46 pm
uw490 wrote:
How do we know to look for the "number that's a multiple of 8, minus 1. Like, 7, 15, 23, etc..." Still not clear why we know to look for this portion of the solution.
J
I guess it just comes with experience with number properties. However, if you notice, the applicability of the 48 even numbers is derived completely from the N and (N+2) term. The (N+1) term isn't even a factor for this part.

Well, you might think to yourself, why is it even here? At this point, you just have to reason out that (a multiple of 8) * (N) * (N+2) will be divisible by 8. Thus, any number such that N+1 is divisible by 8 will work, and that's the 7, 15, 23 etc... part. I know it can be a little tricky to see by yourself, but in these kinds of comprehensive number property questions, it's best to try different combinations of even/odd, positive/negative, and any other way to dichotimize numbers that you can think of if you can't recognize the number properties.
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by uptowngirl92 » Sun Sep 27, 2009 7:16 pm
Amazing explanation but jus one querie..
when we are dealing with the second part..we are dealing only with ODD nos. rite?So when we are counting the nos. divisible by 8 shouldnt we be counting only 6?The even part has already been dealt with...
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by Simon » Sat Nov 13, 2010 9:39 am
For even . odd. even scenario total number of even numbers = 96-2/2 +1 = 47+1 = 48

For odd. even . odd scenario in which the even number is a multiple of 8 are only 11.

for instance 7.8.9...... 15.16.17..........23.24.25..... up to 87.88.89 but there is no twelfth number (i.e) there is no 95.96.97 as 97 is not part of the numbers provided ( 1 to 96 inclusive)

so 48+11 = 49

probability = 49/96 approx = 5\8


Can someone correct me if i'm wrong????
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by Simon » Sat Nov 13, 2010 9:45 am
also we probably need to take out one more number from the even. odd. even

we can have 2.3.4 or 4.5.6 up to 94.95.96 but not 96.97.98

so we only use 47 numbers

so 47 + 11 = 58 / 96 approx 5/8
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by Rezinka » Mon Nov 15, 2010 11:56 pm
Can we do it using elimination ?
This is how I did it :
Writing down first few sets I discovered every alternate set is divisible, so prob > 1/2 or prob = 1/2. Also wherever a multiple of 8 comes in, additional sequences are divisible.
So, prob is slightly greater than 1/2. Now A,B,C are easily eliminated. E is much greater so I chose D.
I am fine with calculations as well, so should I stick to calculations or I can use such eliminations wherever possible ?
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