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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
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Originally posted by RadhaKrishnan on 28 Jan 2012, 04:03.
Last edited by Bunuel on 28 Jan 2012, 05:12, edited 1 time in total.
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Re: Probability
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28 Jan 2012, 05:12
RadhaKrishnan 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 question: divisibleby12probability121561.htmlHope it helps.
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Re: Probability
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29 Jan 2012, 01:17
Bunuel wrote: RadhaKrishnan 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; from here u can also think this way (though it is a lil bit similar) find how many numbers are divisible by 2 or 8 96/2=48 96/8=12 (48+12)/96=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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18 Feb 2012, 08:03
+1 D Other way is analyzing if there is a patron: 1) If n is an even number: n:2, then 2*3*4 = 24 (divisible by 8) n:4, then 4*5*6 = 120 (divisible by 8) n:6, then 6*7*8= again divisible by 8 We have a patron. So, we have 48 even possible values. 2) If n is an odd number: This only can take place when n+1 is multiple of 8. So, we have 12 possible values. Then, \(\frac{(48 + 12)}{96} = \frac{5}{8}\) D
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Re: PS QUESTION
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06 May 2012, 02:22
Any integer n(n+1)(n+2) will be divisible by 8 if n is a multiple of 2. This gives us 48 numbers between 1 and 96. Additionally, all those numbers for which (n+1) is a multiple of 8 are also divisible by 8. This gives us a further 12 numbers. These numbers are all distinct from the first set because the first set had only even numbers and this set has only odd numbers. Therefore probability = (48+12)/96 = 60/96 = 5/8 Option (D)
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Re: If an integer n is to be chosen at random from the integers
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17 Aug 2014, 00:24
goodyear2013 wrote: If n is an integer between 1 and 96 (inclusive), what is the probability that n×(n+1)×(n+2) is divisible by 8?
A. 1/4 B. 1/2 C. 5/8 D. 3/4 E. 7/8 In such problem always follow a simple rule. The divisor is 8. So take number from 1 to 8 as example If n=1, n(n+1)(n+2)= Not divisible by 8 If n=2, n(n+1)(n+2)= yes If n=3, n(n+1)(n+2)=No. Do upto 8 So, 5 out of 8 are divisible. Hence, 5/8 is the answer. This rule is applicable if last number(96 in this case) is divisible by 8. If you had to pick from 198, still you can apply the rule but be careful. But there is one additional step. Hopefully you can find answer in less than 60 seconds



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Re: If an integer n is to be chosen at random from the integers
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22 Nov 2014, 02:00
Here is the dumb way of doing this question, if the smart ones don't strike your head during the test Since we are looking at divisibility by 8, we consider n to be between 1 and 8. We can multiply the result by 12  since 96/8 = 12. Attachment:
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If an integer n is to be chosen at random from the integers
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Re: If an integer n is to be chosen at random from the integers
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17 May 2013, 13:07
there are total 48 numbers in the form n*n+1*n+2 starting from 2,4,6,8,10.....96 for which it is divisible by 8.
Additionally,there are 12 cases if there is 8 or a multiple of 8 that also divides the form n*n+1*n+2 which starts from 7,15,23,31....95.
The above cases both are nonoverlapping, so we can add them.
adding the above two cases 48+12=60 ans=60/96=5/8 which is D.
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Re: Probability
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30 Jan 2012, 09:10
Bunuel wrote: RadhaKrishnan 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 question: divisibleby12probability121561.htmlHope it helps. Bunnel  Thanks for the explanation. I did not take the condition when n+1 is also divisible by 8



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Re: PS QUESTION
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08 May 2012, 01:05
jayaddula wrote: Bunuel wrote: anjolaolu wrote: (PLS CAN SOMEONE KINDLY EXPLAIN WHY THE ANSWER OF THE QUESTION BELOW IS D, THANK YOU)
If an integer n is to be chosen at random from 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 Merging similar topics. Please ask if anything remains unclear. Hi, please correct where i am going wrong with my apporach. between 1 and 96 inclusive, there are 12 multiples of 8 i.e. 8,16,24,32,40,48,54,64,72,80,88,96. so, if n is any of the 12 numbers, then n.(n+1)(n+2) is divisible by 8. now, based on the above listed, the numbers for n+1 can be 7,15,23,31,39,47,53,63,71,79,87,95 and similary for n+2 = 6,14,22,30,38,46,52,62,70,78,86,96. so, a total of 36, hence 36/98 = 3/8 which is incorrect. thanks jay There are more cases for n(n+1)(n+2) to be divisible by 8. Please read the solution above. 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.
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Re: If an integer n is to be chosen at random from the integers
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12 Jul 2015, 05:40
RadhaKrishnan 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 Alternate solution: Look at the first few sets of 3s: 1,2,3 2,3,43,4,5 4,5,65,6,7 6,7,87,8,9 8,9,10 We see that out of the above 8 sets, favorable cases are 5 (in red). Thus the probability is 5/8. This will repeat till we have 88,89,90 making it 9 total patterns. Now consider after 88,89,90, we get 89,90,91 90,91,9291,92,93 92,93,9493,94,95 94,95,9695,96,9796,97,98So we have another 5 favorable out of the remaining 8 sets. Thus we have final 5/8 as the probability. Thus the final probability = {(5/8)*9+(5/8)*1} / (9+1) = 5/8



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Re: PS QUESTION
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07 May 2012, 13:07
Bunuel wrote: anjolaolu wrote: (PLS CAN SOMEONE KINDLY EXPLAIN WHY THE ANSWER OF THE QUESTION BELOW IS D, THANK YOU)
If an integer n is to be chosen at random from 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 Merging similar topics. Please ask if anything remains unclear. Hi, please correct where i am going wrong with my apporach. between 1 and 96 inclusive, there are 12 multiples of 8 i.e. 8,16,24,32,40,48,54,64,72,80,88,96. so, if n is any of the 12 numbers, then n.(n+1)(n+2) is divisible by 8. now, based on the above listed, the numbers for n+1 can be 7,15,23,31,39,47,53,63,71,79,87,95 and similary for n+2 = 6,14,22,30,38,46,52,62,70,78,86,96. so, a total of 36, hence 36/98 = 3/8 which is incorrect. thanks jay



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Re: Probability
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19 Aug 2013, 01:39
Bunuel wrote: RadhaKrishnan 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 question: divisibleby12probability121561.htmlHope it helps. Hi, Can we use this formula for this problem # of multiples of x in the\ range = frac{Last multiple of x in the range  First multiple of x in the range}{x}+1, Which give value of = 12 total outcome is 96, Please explain this with the above formula. Regards, Rrsnathan.



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Re: If an integer n is to be chosen at random from the integers
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04 Sep 2014, 13:42
Bunuel: Please suggest where I went wrong;
Although I understand your solution, I have a doubt on my approach.
I multiplied n(n+1)(n+2) and I got n3+3n2+2n ... raised to the power;
So what I thought was...I need to find the numbers for which the quotient is zero for the above expression....since the last number of the above expression is 2n, it just needs a minimum 4 to be divisible by 8. Moreover a minimum n=4 will make n3 and 3n2 easily divisible by 8; Therefore, I thought that the above expression will be true for all the multiples of 4...there are 24 multiples of 4 from 1 to 96 inclusive, hence my answer comes out to be 24/96=1/4;
I will be really thankful to you if you can tell me where did I go wrong!!
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Re: If an integer n is to be chosen at random from the integers
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05 Sep 2014, 06:05
execnitinsharma wrote: Bunuel: Please suggest where I went wrong;
Although I understand your solution, I have a doubt on my approach.
I multiplied n(n+1)(n+2) and I got n3+3n2+2n ... raised to the power;
So what I thought was...I need to find the numbers for which the quotient is zero for the above expression....since the last number of the above expression is 2n, it just needs a minimum 4 to be divisible by 8. Moreover a minimum n=4 will make n3 and 3n2 easily divisible by 8; Therefore, I thought that the above expression will be true for all the multiples of 4...there are 24 multiples of 4 from 1 to 96 inclusive, hence my answer comes out to be 24/96=1/4;
I will be really thankful to you if you can tell me where did I go wrong!!
Thanks. You are missing cases there. Won't n^3+3n^2+2n when n is any even number? Also it can be divisible by 8, even if individual terms are not but the sum is. For example check for n =7.
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Re: If an integer n is to be chosen at random from the integers
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16 Jul 2016, 02:41
megha_2709 wrote: Bunuel wrote: RadhaKrishnan 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 question: divisibleby12probability121561.htmlHope it helps. Hi, Thank you for posting such a good explanation , however I could not understand how can you categorize the numbers in group of 8 . Probability is Fav/Total . Shouldn't we consider all 96 values and find out how many are satisfying our conditions , cant understand how you arrive at 5/8. If you can please explain. Sorry if this sounds too basic. Regards Megha 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
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Re: If an integer n is to be chosen at random from the integers
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16 Nov 2016, 08:55
RadhaKrishnan 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 We are given that an integer n is to be chosen at random from the integers 1 to 96 inclusive, and we need to determine the probability that n(n + 1)(n + 2) will be divisible by 8. We should recall that when a number is divisible by 8, it is divisible by 2^3, i.e., three factors of 2. We should also recognize that n(n + 1)(n + 2) is the product of three consecutive integers. Case 1: n is even. Any time that n is even, n + 2 will also be even. Moreover, either n or n + 2 will be divisible by 4, and thus n(n + 1)(n + 2) will contain three factors of 2 and will be divisible by 8. Since there are 96 integers between 1 and 96, inclusive, and half of those integers are even, there are 48 even integers (i.e., 2, 4, 6, …, 96) from 1 to 96 inclusive. Thus, when n is even, there are 48 instances in which n(n + 1)(n + 2) will be divisible by 8. Case 2: n is odd. If n is odd, then n(n + 1)(n + 2) still can be divisible by 8 if the factor (n + 1) is a multiple of 8. So, let’s determine the number of multiples of 8 between 1 and 96 inclusive. Number of multiples of 8 = (96  8)/8 + 1 = 88/8 + 1 = 12. Thus, when n is odd, there are 12 instances in which n(n + 1)(n + 2) will be divisible by 8. In total, there are 48 + 12 = 60 outcomes in which n(n + 1)(n + 2) will be divisible by 8. Thus, the probability that n(n + 1)(n + 2) is divisible by 8 is: 60/96 = 10/16 = 5/8. Answer: D
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Re: If an integer n is to be chosen at random from the integers
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07 Feb 2017, 07:48
My approach was similar to Bunuel's and AdmitJA's, but I wanted to offer it because it was simple and quick. I was able to do this one in just over a minute.
When I read the question, I paid attention to two things right away. The first was that \((n)(n+1)(n+2)\) is the product of three consecutive integers. Because \(8=2^3\), I knew that I needed to find three 2s in the factorization of whatever 3 consecutive numbers I used. The second thing that I noticed is that 96 is divisible by 8. This indicated to me that whatever pattern I noticed in the first 8 numbers would be repeated 12 times through 96. Since I recognized that the pattern would be repeated, I knew that I only needed to look at the first 8 numbers.
I listed the numbers out: 1 2 3 4 5 6 7 8.
Since 8 has three 2s in its factorization, I knew that n=6,7,8 would all be divisible by 8. I saw that n=2 would work because 2 has one 2 and 4 has two 2s. Similarly, I saw that n=4 would work because 4 has two 2s and 6 has one 2. That gives us 5 options in the first 8 numbers. So the answer is 5/8.
As mentioned above, this ratio will be the same for every 8 numbers, so 5/8 will be true of 8x where x is any positive integer. A trickier version of this question would have been to make the number NOT divisible by 8. In that case, I think you should still find the pattern for every 8 numbers, but you'd also want to look at the "extra" numbers to figure out the fraction.
Take 53, for instance. You'd want to recognize that 8 goes into 53 six times with a remainder of 5. This means that you'd have the \(5/8\) ratio for 6 sets of 8 but also an "incomplete" set of n = 1,2,3,4,5. In the first 48 numbers, you'd have 30 that would be divisible by 8. In the "incomplete" set you'd have 2 (since n=2 and n=4 are both divisible by 8). Thus you'd have 32/53 numbers 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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15 May 2018, 10:26
siddreal wrote: RadhaKrishnan 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 VeritasPrepKarishma Pls show us your approach to this question. Though I understood the solutions mentioned in the thread, I am finding it difficult to believe that I might think of this approach during actual exam. What I would do is pretty much what Bunuel has done in his solution. Note that the moment I see n(n + 1)(n + 2), I think of divisibility in a bunch of consecutive numbers. In 3 consecutive integers, if n is even, (n+2) is even too. If there are two consecutive even integers, one of them will be a multiple of 4. So if n is even, (n+2) is even too and one of them is definitely a multiple of 4. So n(n + 1)(n + 2) becomes divisible by 8 in each case that n is even. From 1 to 96, half the cases have even n so this mean 48 cases. Alternatively, when n is odd, n+1 is even. But then n+2 is odd too. So to be a multiple of 8, (n+1) will need to be a multiple of 8. Hence this gives us another 12 cases (n+1 goes from 8 to 96). Note that there will be no overlap in the two since here n is definitely odd. Total we have 60 cases of the possible 96 which gives 60/96 = 5/8 For these properties of numbers, see https://www.veritasprep.com/blog/2011/0 ... cormath/https://www.veritasprep.com/blog/2011/0 ... hpartii/
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