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If an integer n is to be chosen at random from the integers

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If an integer n is to be chosen at random from the integers [#permalink] New post 02 Feb 2005, 09:54
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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: if-an-integer-n-is-to-be-chosen-at-random-from-the-integers-126654.html
[Reveal] Spoiler: OA
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 [#permalink] New post 03 Feb 2005, 05:29
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
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 [#permalink] New post 03 Feb 2005, 05:54
what about 6 7 8 or 7 8 9 or 8 9 10 ? it is a different pattern !
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 [#permalink] New post 03 Feb 2005, 05:56
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
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 [#permalink] New post 03 Feb 2005, 08:13
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, 10:37, edited 1 time in total.
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 [#permalink] New post 03 Feb 2005, 08:30
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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] New post 03 Feb 2005, 08:41
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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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 [#permalink] New post 03 Feb 2005, 10:47
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
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 [#permalink] New post 03 Feb 2005, 12:04
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.
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 [#permalink] New post 03 Feb 2005, 16:50
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, 19:02, edited 1 time in total.
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 [#permalink] New post 03 Feb 2005, 17:37
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?
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 [#permalink] New post 03 Feb 2005, 20:14
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.
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 [#permalink] New post 03 Feb 2005, 20:41
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 = 2P-1
=4(2P-1)(2P-2 +1) (2P-1 +1)
=4(2P-1)(2P-1)(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 = 2K-1:
=(2K-1)(2K)(2K+1)
=2(2K-1)(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] New post 17 Oct 2013, 13:46
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post 17 Oct 2013, 23:40
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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: {1-8}, {9-16}, {17-24}, ..., {89-96} 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: if-an-integer-n-is-to-be-chosen-at-random-from-the-integers-126654.html
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Re: If an integer n is to be chosen at random from the integers   [#permalink] 17 Oct 2013, 23:40
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