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If an integer n is to be chosen at random from the integers [#permalink]
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29 Dec 2009, 05:41
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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. \(\frac{1}{4}\) B. \(\frac{3}{8}\) C. \(\frac{1}{2}\) D. \(\frac{5}{8}\) E. \(\frac{3}{4}\) OPEN DISCUSSION OF THIS QUESTION IS HERE: ifanintegernistobechosenatrandomfromtheintegers126654.html
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Last edited by Bunuel on 20 Apr 2015, 08:16, edited 3 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 [#permalink]
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29 Dec 2009, 09:13
kirankp 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. 41 B. 83 C. 21 D. 85 E. 43 are you sure the options are correct?? ...as per my understanding all even numbers in this range 2 4 6 8 till 96 will be divisible by 8 for n(n+1)(n+2)....what am I missing here



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Re: If an integer n is to be chosen at random from the integers [#permalink]
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29 Dec 2009, 09:28
i am sorry, the options were wrong, have edited it



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Re: If an integer n is to be chosen at random from the integers [#permalink]
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29 Dec 2009, 09:31
kirankp wrote: i am sorry, the options were wrong, have edited it hmmm...will go with C1/2



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Re: If an integer n is to be chosen at random from the integers [#permalink]
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29 Dec 2009, 09:50
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. 41 B. 83 C. 21 D. 85 E. 43 \(n(n+1)(n+2)\) is divisible by 8 in two cases: 1. When \(n\) is even: \(n=2k\) > \(n(n+1)(n+2)=2k(2k+1)(2k+2)=4k(2k+1)(k+1)\) > either \(k\) or \(k+1\) is even so 8 is a multiple of \(n(n+1)(n+2)\). # of even numbers (between 1 and 96)=48 AND 2. When \(n+1\) is divisible by 8. > \(n+1=8p\) (\(p\geq{1}\)), \(n=8p1\) > \(8p1\leq{96}\) > \(p\leq{12.1}\) > 12 such numbers. Also note 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. Total=48+12=60 Probability: \(\frac{60}{96}=\frac{5}{8}\) Answer: D.
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Re: If an integer n is to be chosen at random from the integers [#permalink]
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29 Dec 2009, 09:52
Bunuel wrote: n(n+1)(n+2) is divisible by 8 in two cases:
1. When n is even: No of even numbers (between 1 and 96)=48
AND
2. When n+1 is divisible by 8. > n+1=8p, n=8p1 > 8p1<=96 > p<=12.1 > 12 such numbers
Also note that these two sets have no overlaps.
Total=48+12=60
Probability=60/96=5/8
Answer: D. oops



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Re: If an integer n is to be chosen at random from the integers [#permalink]
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02 Jan 2010, 03:53
agree with Bunuel here. if n=even, then n(n+2) will always have at least three 2's as factors (2x2x2=8), therefore 8 will always be a factor, where n= even. there are 48.
if (n+1)= divisible by 8, then the whole equation will also be divisible by 8: n=7,15,23,31,39,47,55,63,71,79,87,95 = 12 possibilities.
48+12=60  number of times expression will be divisible by 8. 96= number of possibilities.
probability = 60/96 = 5/8



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Re: If an integer n is to be chosen at random from the integers [#permalink]
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02 May 2010, 09:16
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
Last edited by chaitu1315 on 02 May 2010, 09:24, edited 1 time in total.



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Re: If an integer n is to be chosen at random from the integers [#permalink]
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02 May 2010, 09:22
chaitu1315 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 IMO A This is only possible when n is multiple of 4. now from 4 to 96 there are 24 such numbers, thus probability = 24/96 = 1/4
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Re: If an integer n is to be chosen at random from the integers [#permalink]
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02 May 2010, 11:47
gurpreetsingh wrote: chaitu1315 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 IMO A This is only possible when n is multiple of 4. now from 4 to 96 there are 24 such numbers, thus probability = 24/96 = 1/4 Gurpreet you did not consider the options mentioned below ..! What about when n =6 or may be 15 then also the equation is divisible by 8 and hence we need to consider them as well.. so in my opinion the expression will be divisible by 8 whenever n is divisible by 4 or (n+ 2) is divisible by 4 or when (n+1) is divisible by 8 1st Option when n is divisible by = total numbers = 24 ( as calculated by Gurpreet above 96/4= 24) 2nd option when n +1 is divisible by 8 = 12 ( 96/8 ) 3rd Option when n+2 is divisible by 4 = 24 ( 96/4) in total = 24 + 24 + 12 = 60 the probability = 60/ 96 = 5/ 8 D



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Re: If an integer n is to be chosen at random from the integers [#permalink]
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02 May 2010, 14:46
yes right, i missed many.
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Re: If an integer n is to be chosen at random from the integers [#permalink]
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kirankp 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. \(\frac{1}{4}\) B. \(\frac{3}{8}\) C. \(\frac{1}{2}\) D.\(\frac{5}{8}\) E. \(\frac{3}{4}\) IMO 'D' Consider the following possibilities: 1 2 3 2 3 4 3 4 5 4 5 6 5 6 7 6 7 87 8 98 9 10....... ...... 96 97 98 If you notice every alternate pair is divisible by 8 => 48 such pairs If you notice the pair(marked in blue) it has '8' in it and hence divisible by '8'=>12 such pairs =>prob=48+12 / 96 = 5/8
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Re: If an integer n is to be chosen at random from the integers [#permalink]
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11 May 2010, 10:15
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 The problem can be solved in a crude way, but is there a better alternative? Thanks



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Re: If an integer n is to be chosen at random from the integers [#permalink]
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11 May 2010, 19:52
I do not know how to solve this. Where did you get this problem?



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Re: If an integer n is to be chosen at random from the integers [#permalink]
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11 May 2010, 20:09
ajitsah 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 The problem can be solved in a crude way, but is there a better alternative? Thanks Probability = required cases/ total cases. required cases . Case 1. n(n + 1)(n + 2) will be divisible by 8 when n is even, in that case, n will be multiple of 2 and n+2 of 4. so total even numbers between 1 and 96 = 48 Case 2. when n+1 = 8k form => n = 8k1 such numbers from 1 to 96 are 12... it starts from 7 to 95 which diff of 8.This forms an AP series. So required cases = 48+12 = 60 total cases = 96 probability = 60/96 = 5/8 Hence D
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Re: If an integer n is to be chosen at random from the integers [#permalink]
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17 May 2010, 15:30
good problem great explanation.



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Re: If an integer n is to be chosen at random from the integers [#permalink]
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11 Aug 2010, 06:36
I am curious... how should we approach this type of question...? for eg this question says 8 if question says 11 or may be 12 do we test numbers? what if there are certain exception in middle? (in case of 8 it seems it goes continously...)
do we remember for each numbers?



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Re: If an integer n is to be chosen at random from the integers [#permalink]
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08 Sep 2010, 11:18
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. 3/8
Last edited by udaymathapati on 09 Sep 2010, 05:47, edited 1 time in total.



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Re: If an integer n is to be chosen at random from the integers [#permalink]
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20 Apr 2015, 08:16
kirankp 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. \(\frac{1}{4}\) B. \(\frac{3}{8}\) C. \(\frac{1}{2}\) D. \(\frac{5}{8}\) E. \(\frac{3}{4}\) \(n(n + 1)(n + 2)\) 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 question: divisibleby12probability121561.htmlOPEN DISCUSSION OF THIS QUESTION IS HERE: ifanintegernistobechosenatrandomfromtheintegers126654.html
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