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If an integer n is to be chosen at random from the integers [#permalink]
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02 Feb 2005, 10:54
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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 OPEN DISCUSSION OF THIS QUESTION IS HERE: ifanintegernistobechosenatrandomfromtheintegers126654.html
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Director
Joined: 07 Jun 2004
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when ever we have E O E we get divisible by 8 eg 2 3 4, 10 11 12 when wen ever we have O E O no divisible by 8
Answer = 1/2



VP
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what about 6 7 8 or 7 8 9 or 8 9 10 ? it is a different pattern !



Director
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dude u r right i did not realize that 7 8 9 even though O E O is dvisible by 8
how did u get 3/ 8



VP
Joined: 18 Nov 2004
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I actually get 7/12.....every 24 numbers have 14 Ns that will work....so we have 14*4 = 56 numbers
p(e) = 56/96 = 7/12...not in the choices
EDITED:
every 24 numbers actually has 15 numbers that will work, not 14.....so we have 60 numbers in total
p(e) = 60/96 = 5/8
Last edited by banerjeea_98 on 03 Feb 2005, 11:37, edited 1 time in total.



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if n is even n(n+1)(n+2) is necessarily a multiple of 8 : P = 1/2
if n is odd then n(n+1)(n+2) is a multiple of 8 if only (n+1) is.
so n= from 7 to 95 : 12 numbers P=12/96=1/8
So P=1/2+1/8= 5/8
D it is.



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PS divisible by 8 [#permalink]
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03 Feb 2005, 09:41
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I get 5/8 as well
1 to 96 inclusive means we have 48 odd and 48 even integers
E O E / 8 = Integer, therefore we have 48 / 96 numbers divisible by 8
O E O / 8 = Not Integer
We cannot forget multiples of 8 from 1 to 96
We have 12 numbers that are multiple of 8
Therefore, 48/96 + 12/96 = 60/96 = 5/8
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twixt wrote: if n is even n(n+1)(n+2) is necessarily a multiple of 8 : P = 1/2
if n is odd then n(n+1)(n+2) is a multiple of 8 if only (n+1) is. so n= from 7 to 95 : 12 numbers P=12/96=1/8
So P=1/2+1/8= 5/8
D it is.
yaa, that's my reasoning.
good explanation



Manager
Joined: 01 Jan 2005
Posts: 166
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I know i am late. The answer will be 5/8.
Every 8 nos there will be 5 such combinations. Thus 96/8=12
12*5 =60 such combinations.
Probability=60/96=5/8.



Director
Joined: 19 Nov 2004
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Can somebody give a proof for the following statements Algebraically:
1) if n is even, n(n+1)(n+2) is necessarily a multiple of 8 when n>=2
2) if n is odd, then n(n+1)(n+2) is a multiple of 8 if only (n+1) is.
Twixt?
Last edited by nocilis on 03 Feb 2005, 20:02, edited 1 time in total.



Director
Joined: 27 Dec 2004
Posts: 898

banerjeea_98 wrote: I actually get 7/12.....every 24 numbers have 14 Ns that will work....so we have 14*4 = 56 numbers p(e) = 56/96 = 7/12...not in the choices EDITED: every 24 numbers actually has 15 numbers that will work, not 14.....so we have 60 numbers in total p(e) = 60/96 = 5/8
Banerjeea, your second approach is quite simple however, how did you figure that there are 15 numbers that will work in every 24 numbers?



SVP
Joined: 03 Jan 2005
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nocilis wrote: Can somebody give a proof for the following statements Algebraically:
1) if n is even, n(n+1)(n+2) is necessarily a multiple of 8 when n>=2 2) if n is odd, then n(n+1)(n+2) is a multiple of 8 if only (n+1) is.
Twixt?
Look at my post a couple posts above, then you'll get the idea. Use 2k and 2k+1 for even and odd.



Director
Joined: 19 Nov 2004
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Here is the proof
1) if n is even, n(n+1)(n+2) is necessarily a multiple of 8 when n>=2
n = 2K:
= 2K(2K+1)(2K+2)
= 4K(2K+1)(K+1)
Using the above, all I can say is it is a multiple of 4.
Let us proceed further..
K even:
If K is even, we can tell that above is a multiple of 8.
K Odd:
If K is odd = 2P1
=4(2P1)(2P2 +1) (2P1 +1)
=4(2P1)(2P1)(2p)
Aha ... it is a multiple of 8!
Proved!
2) if n is odd, then n(n+1)(n+2) is a multiple of 8 if only (n+1) is.
n = 2K1:
=(2K1)(2K)(2K+1)
=2(2K1)(K)(2K+1)
Using the above, all I can say is it is a multiple of 2.
Although, we can say that the above product is a multiple of 8 if n is odd only if n+1 is a multiple of 8 looking at the product as n and n+2 will still be odd.
Proved.



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Re: If an integer n is to be chosen at random from the integers [#permalink]
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17 Oct 2013, 14:46
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Re: If an integer n is to be chosen at random from the integers [#permalink]
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18 Oct 2013, 00:40
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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 \(n(n + 1)(n + 2)\) is divisible by 8 in two cases: A. \(n=even\), in this case \(n+2=even\) too and as \(n\) and \(n+2\) are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8; B. \(n+1\) is itself divisible by 8; (Notice that these two sets have no overlaps, as when \(n\) and \(n+2\) are even then \(n+1\) is odd and when \(n+1\) is divisible by 8 (so even) then \(n\) and \(n+2\) are odd.) Now, in EACH following groups of 8 numbers: {18}, {916}, {1724}, ..., {8996} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8. Answer: D. OPEN DISCUSSION OF THIS QUESTION IS HERE: ifanintegernistobechosenatrandomfromtheintegers126654.html
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Re: If an integer n is to be chosen at random from the integers
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