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# N is from 1 to 96 inclusive, what is the probability of N

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CEO
Joined: 15 Aug 2003
Posts: 3467
Followers: 61

Kudos [?]: 718 [0], given: 781

N is from 1 to 96 inclusive, what is the probability of N [#permalink]  08 Oct 2003, 00:23
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N is from 1 to 96 inclusive, what is the probability of N such that N(N+1)(N+2)is evenly divisible by

problem #1 . Eight
problem #2 . Four
problem #3 . Three

#1
EIGHT:
All "sets" of even N's are candidates
Example 2,3,4... 4,5,6 ... 6 7 8
There are 48 such "Sets"
Total possible sets = 96 (one with each N)

Probability = 48/96 =1/2 .... Is this correct.

With # 3 , i find that a set of 3 consecutive positive integers is always divisible by 3...is that right?

Thanks
praetorian
Intern
Joined: 29 Aug 2003
Posts: 49
Location: Detroit, MI
Followers: 0

Kudos [?]: 2 [0], given: 0

Re: PS : Probability...consecutive integers [#permalink]  08 Oct 2003, 11:51
praetorian123 wrote:
N is from 1 to 96 inclusive, what is the probability of N such that N(N+1)(N+2)is evenly divisible by

problem #1 . Eight
problem #2 . Four
problem #3 . Three

#1
EIGHT:
All "sets" of even N's are candidates
Example 2,3,4... 4,5,6 ... 6 7 8
There are 48 such "Sets"
Total possible sets = 96 (one with each N)

Probability = 48/96 =1/2 .... Is this correct.

With # 3 , i find that a set of 3 consecutive positive integers is always divisible by 3...is that right?

Thanks
praetorian

#3 - Every 3rd number is a multiple of 3. So, one of N, (N+1), (N+2) is always divisible by 3. So, you are right. Every N satisfies that criteria. For this case, prob. should be 1.

#2 - Taking the least possible number for N i.e 1, N*(N+1)*(N+2) will be 6. For N <= 6, the following theory should hold.

As long as one of the numbers N, N+1, N+2 is divisible by N, the whole number i.e. N*(N+1)*(N+2) will be divisible by N.

I think one can follow this logic to see that 24 numbers between 1 and 96 are divisible by 4 and each such number can occur in 3 positions - N or (N+1) or (N+2) in this multiplication expression.

So, answer should be (24*3)/96 = 3/4

Amar.
CEO
Joined: 15 Aug 2003
Posts: 3467
Followers: 61

Kudos [?]: 718 [0], given: 781

Re: PS : Probability...consecutive integers [#permalink]  08 Oct 2003, 16:40
[quote="amarsesh]
#3 - Every 3rd number is a multiple of 3. So, one of N, (N+1), (N+2) is always divisible by 3. So, you are right. Every N satisfies that criteria. For this case, prob. should be 1.

#2 - Taking the least possible number for N i.e 1, N*(N+1)*(N+2) will be 6. For N <= 6, the following theory should hold.

As long as one of the numbers N, N+1, N+2 is divisible by N, the whole number i.e. N*(N+1)*(N+2) will be divisible by N.

I think one can follow this logic to see that 24 numbers between 1 and 96 are divisible by 4 and each such number can occur in 3 positions - N or (N+1) or (N+2) in this multiplication expression.

So, answer should be (24*3)/96 = 3/4

Amar.[/quote]

Thanks!
Great Explanation Amar !

for 5, there are 19 #'s and any of these numbers can hold 3 positions.... 19*3 / 96 = 19/32 ...correct??

for 6 , there are 16 #'s = > 16*3/96 =1/2 ..correct??

Stolyar/Amar/Vicks,
Could you share any other properties like these with us?

Thanks
Praetorian
Re: PS : Probability...consecutive integers   [#permalink] 08 Oct 2003, 16:40
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