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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 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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Originally posted by plaza202 on 11 Nov 2007, 21:09.
Last edited by Bunuel on 26 Jul 2014, 09:24, edited 2 times in total.
Renamed the topic, edited the question and added the OA.



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Re: If an integer n is to be chosen at random from the integers
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11 Nov 2007, 21:22
plaza202 wrote: If an integer n is to be chosen at randon 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.
please help .................
I will go with D 5/8
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?
total nos of ways in which we can choose n = 96
n(n + 1)(n + 2) will be divisible by 8?
case 1: n = odd then n+2 =odd & n+1 will be even i.e this needs get divided by 8, hence is a multiple of 8 so we have 8..96 = 12 multiples to fill the n+1 pos hence 12 ways
case 2: n is even then n+2 will be even & the product will be divisible by 24 & thus 8
so nos of values that can be used for n= 2....96 (all even nos) i.e 48 nos
total = 48+12 =60 ways
so reqd P =60/96 =5/8



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Re: If an integer n is to be chosen at random from the integers
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12 Nov 2007, 10:57
The question is asking if you select n what is the probability that the the product of n and the next two consecutive numbers is divisible by 8. For that to happen n would have to be an even number because if n is even then the third number would have to be an even number greater than 2.
ex. 2,3,4
The prime roots of 2 are of course 2 and the prime roots of 4 are 2 and 2. So any 3 consecutive numbers starting with an even number will be divisible by 8(2*2*2).
There are 48 even numbers and 96 possibilites. So 48/96 = 1/2.
Ans C



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Re: If an integer n is to be chosen at random from the integers
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12 Nov 2007, 11:01
gixxer1000 wrote: The question is asking if you select n what is the probability that the the product of n and the next two consecutive numbers is divisible by 8. For that to happen n would have to be an even number because if n is even then the third number would have to be an even number greater than 2.
ex. 2,3,4
The prime roots of 2 are of course 2 and the prime roots of 4 are 2 and 2. So any 3 consecutive numbers starting with an even number will be divisible by 8(2*2*2).
There are 48 even numbers and 96 possibilites. So 48/96 = 1/2.
Ans C
What abt 7,8,9
starts with an odd integer but is still divisible by 8



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Re: If an integer n is to be chosen at random from the integers
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12 Nov 2007, 13:00



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Re: If an integer n is to be chosen at random from the integers
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26 Aug 2008, 14:27
5/8 should be the answer. 48 numbers are even for n 12 triplets have multiples of 8 in them  this implies that those odd numbers that are in those triplets are all game, e.g., 789 ,151617, 232425 etc. Therefore, 7, 15, 23 would also be counted as possible n. There are 12 such nos. Therefore, the total should be 48 +12 = 60. Hence, prob. = 60/96 = 5/8



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Re: If an integer n is to be chosen at random from the integers
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28 Sep 2009, 05:06
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.
Soln:
n(n+1)(n+2) will be divisible by 8 when n is a multiple of 2 or when (n+1) is a multiple of 8.
Thus when n is even, this whole expression will be divisible by 8. from 1 to 96, there are 48 even integers.
Now when (n+1) is multiple by 8, we have 12 such values for (n+1)
probability that n(n+1)(n+2) will be divisible by 8 = (48 + 12)/96 = 60/96 = 5/8
Ans is D



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Re: If an integer n is to be chosen at random from the integers
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02 May 2011, 23:13
choosing the first 8 numbers as sample period (1,2,3),(2,3,4),(3,4,5),(4,5,6),(5,6,7),(6,7,8),(7,8,9),(8,9,10) for n= 1 to 8 respectively. 5 of the sets are divisible by 8 hence 5/8.
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If an integer n is to be chosen at random from the integers
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26 Jul 2014, 09:25
plaza202 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 \(n(n + 1)(n + 2)\) 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. Similar questions to practice: ifintegercisrandomlyselectedfrom20to99inclusive121561.htmlanintegernbetween1and99inclusiveistobechosenat160998.htmlOPEN 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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