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If an integer n is to be chosen at random from the integers [#permalink]
 28 Jan 2012, 04:03
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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
Last edited by Bunuel on 28 Jan 2012, 05:12, edited 1 time in total.
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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: {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. Similar question: divisible-by-12-probability-121561.htmlHope it helps.
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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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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: {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. Similar question: divisible-by-12-probability-121561.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: If an integer n is to be chosen at random from the integers [#permalink]
18 Feb 2012, 08:03
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+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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(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
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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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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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LalaB 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; 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 hello please why do you divide by 2? thanks best regards
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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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If an integer n is to be chosen at random [#permalink]
19 Feb 2013, 02:15
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+) will be divisible by8?
(A)1/4
(B)3/8
(C)1/2
(D)5/8
(E)3/4
Need explanation.............
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Re: If an integer n is to be chosen at random [#permalink]
19 Feb 2013, 03:25
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Re: If an integer n is to be chosen at random from the integers [#permalink]
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 non-overlapping, 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: If an integer n is to be chosen at random from the integers
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17 May 2013, 13:07
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