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integers and prob [#permalink]
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19 Jan 2008, 06:19
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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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Re: integers and prob [#permalink]
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19 Jan 2008, 13:18
Btake a look at integer sequence: 1,2,3,4,5,6,7, 8,9,10,11,12,13,14,15, 16,17,18,19 to catch 8*n we have to start with 8n2, 8n1, 8n 1,2,3,4,5, [6,7,8],9,10,11,12,13,14,15, 16,17,18,19 1,2,3,4,5,6, [7,8,9],10,11,12,13,14,15, 16,17,18,19 1,2,3,4,5,6,7, [8,9,10],11,12,13,14,15, 16,17,18,19 Therefore, we have 3 combinations for each 8 consecutive integers. \(p=\frac38\)
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Re: integers and prob [#permalink]
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19 Jan 2008, 13:33
walker wrote: B
take a look at integer sequence:
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19
to catch 8*n we have to start with 8n2, 8n1, 8n
1,2,3,4,5,[6,7,8],9,10,11,12,13,14,15,16,17,18,19 1,2,3,4,5,6,[7,8,9],10,11,12,13,14,15,16,17,18,19 1,2,3,4,5,6,7,[8,9,10],11,12,13,14,15,16,17,18,19
Therefore, we have 3 combinations for each 8 consecutive integers. \(p=\frac38\) Why do we have to start at 8? if n=2 then 2(3)(4) is divisible by 8? I get every even number from 196 when n(n+1)(n+2) is divisible by 8.



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Re: integers and prob [#permalink]
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19 Jan 2008, 13:41
ah... you are right Dtake a look at integer sequence: 1,2,3,4,5,6,7, 8,9,10,11,12,13,14,15, 16,17,18,19 to catch 8*n we have to start with 8n2, 8n1, 8n 1,2,3,4,5, [6,7,8],9,10,11,12,13,14,15, 16,17,18,19 1,2,3,4,5,6, [7,8,9],10,11,12,13,14,15, 16,17,18,19 1,2,3,4,5,6,7, [8,9,10],11,12,13,14,15, 16,17,18,19 Therefore, we have 3 combinations for each 8 consecutive integers. to catch 8*n we also have to start with other even integers 8n6, 8n4 1, [2,3,4],5,6,7, 8,9,10,11,12,13,14,15, 16,17,18,19 1,2,3, [4,5,6],7, 8,9,10,11,12,13,14,15, 16,17,18,19 \(p=\frac38+\frac28=\frac58\)
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Re: integers and prob [#permalink]
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19 Jan 2008, 13:43
GMATBLACKBELT wrote: I get every even number from 196 when n(n+1)(n+2) is divisible by 8. plus integers like (7,8,9) or (15,16,17)....so 4/8+1/8=5/8
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Re: integers and prob [#permalink]
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19 Jan 2008, 23:01
marcodonzelli 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 1,2,3,...96. n can be chosen in 96 ways. ( This is total number of samples ) \(n*(n+1)*(n+2)/8\) implies we should find number of favorable samples in which we have \(n*(n+1)*(n+2) = 8(min), 16, 32, ...96(max)\) \(8 = 2^3\) ( Thus total number of divisors of 8 will be \((3+1)=4\) ) which are 1, 2, 4, 8. From this we get 3 different possibilities; \(2*(2+1)*(2+2)\) \(4*(4+1)*(4+2)\) \(8*(8+1)*(8+2)\) But with 8 we can also have; \(6*(6+1)*(6+2)\) \(7*(7+1)*(7+2)\) Similarly for \(16 = 2^4\) ( Thus total number of divisors of 8 will be \((4+1)=5\) ) which are 1, 2, 4, 8, 16. From this also we get 4 different possibilities; \(2*(2+1)*(2+2)\)\( 4*(4+1)*(4+2)\) \(8*(8+1)*(8+2)\) Two more associated with it as shown above. \(16(16+1)*(16+2)\) Two more associated with it depending on the position of 16. Two of them are already in the sequence, so 3 different. and so on... with each multiple of 8, we will get 3 different possibilities. Now, how many are the total number of number which are multiple of 8 from 8...96. \((968/8)+1=12\) Each of these 12 different multiples have 3 different possibilities in which 8 can be the divisor. Thus total number of favorable outcomes = \(12*3 + 2*(2+1)*(2+2) + 4*(4+1)*(4+2) = 36+1+1\) \(P = 38/96\) Did I miss something????



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Re: integers and prob [#permalink]
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24 Jan 2008, 11:43
C 1/2 is my answer. A no can be divided by eight if we can divide that no by both 2 and 4. n(n+1)(n+2) , three consecutive nos can always be divided by 2, and if n is even, we can divide by 4 also Therefore for all even n we can divide n(n+1)(n+2) by 8 we have 48 even and 96 total nos so, prob = 48/96= 1/2 whats the OA pls?????



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Re: integers and prob [#permalink]
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24 Jan 2008, 11:49
Dellin wrote: C 1/2 is my answer. A no can be divided by eight if we can divide that no by both 2 and 4. n(n+1)(n+2) , three consecutive nos can always be divided by 2, and if n is even, we can divide by 4 also Therefore for all even n we can divide n(n+1)(n+2) by 8 we have 48 even and 96 total nos so, prob = 48/96= 1/2
read above. plus integers like (7,8,9) or (15,16,17)....so 1/2+1/8=5/8
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Re: integers and prob [#permalink]
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20 Feb 2008, 01:46
walker wrote: Dellin wrote: C 1/2 is my answer. A no can be divided by eight if we can divide that no by both 2 and 4. n(n+1)(n+2) , three consecutive nos can always be divided by 2, and if n is even, we can divide by 4 also Therefore for all even n we can divide n(n+1)(n+2) by 8 we have 48 even and 96 total nos so, prob = 48/96= 1/2
read above. plus integers like (7,8,9) or (15,16,17)....so 1/2+1/8=5/8 walker can you help me with this approach? 96=2^5 * 3=2^3 * 2^2 *3= so we have a total of 12 factors. to find multiples of 8 we must find all multiples of 2^3, so 6 factors, 1/2. what do I miss?



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Re: integers and prob [#permalink]
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20 Feb 2008, 02:35
marcodonzelli wrote: walker can you help me with this approach?
96=2^5 * 3=2^3 * 2^2 *3= so we have a total of 12 factors. to find multiples of 8 we must find all multiples of 2^3, so 6 factors, 1/2. what do I miss? I think the approach is not correct. You missed the fact that we have 3 consecutive integers. For example, 4,5,6 are not multiplies of 8 but 4*5*6 is divisible by 8.
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Re: integers and prob [#permalink]
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20 Feb 2008, 10:20
marcodonzelli 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 Total numbers 8*12 There are 12 numbers divisible by 8 > 3*12 (if 8 is an example  (6,7,8), (7,8,9), (8,9,10)) and 12 numbers divisible by 4 but not divisible by 8 > 2*12 (if 4 is an example (2,3,4) and (4,5,6)) The answer 5/8 > D



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Re: integers and prob [#permalink]
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20 Feb 2008, 12:24
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My 30 second answer:
if n is even, then n*n+2 is divisible by 8. so we have boom 50% so far.
then, we consider if n+1 is divislbe by 8, which is another 12 times
so 48 + 12 / 96 = 60/96 = 30/48 = 5/8
QED



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Re: integers and prob [#permalink]
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20 Feb 2008, 23:56
StartupAddict wrote: My 30 second answer:
if n is even, then n*n+2 is divisible by 8. so we have boom 50% so far.
then, we consider if n+1 is divislbe by 8, which is another 12 times
so 48 + 12 / 96 = 60/96 = 30/48 = 5/8
QED that's a great job really. quick and easy. kudo for you



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Re: integers and prob [#permalink]
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17 Mar 2008, 03:32
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marcodonzelli wrote: StartupAddict wrote: My 30 second answer:
if n is even, then n*n+2 is divisible by 8. so we have boom 50% so far.
then, we consider if n+1 is divislbe by 8, which is another 12 times
so 48 + 12 / 96 = 60/96 = 30/48 = 5/8
QED that's a great job really. quick and easy. kudo for you I didn't imagine it could be so easy, without any use of calculus and without trying with numbers, somewhat very difficult to do in 2 mins during the exam: 96/8=12 numbers divisible by 8 (12 + 12 + 12)/96=3/8. it took me 1 second



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Re: integers and prob [#permalink]
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28 Apr 2008, 01:09
OK, here is a long explanation for a problem that took me less than a minute. First and most important, we know that for n(n+1)(n+2) to be divisible by 8, either n or n+1 or n+2 must be divisible by 8 Second we have to realize that we are dealing with consecutive integers. We also realize that as we have consecutive integers and multiplication, every even integer n from 2 to 96 satisfy n(n+1)(n+2) to be divisible by 8, allowing us to quickly count all even integers from 2 to 96 to be 47, when we look when n or n+1 or n+2 is divisible by 8. Realizing the above three in a snap, we simply check when n or n+1 or n+2 is divisible by 8: 1. n is divisible by 8 when n is 8,16,32,36,40,48, 56,64,72,80,88,96, but these even integers are already included in our conclusion that every even integer n from 2 to 96 satisfy the condition that n(n+1)(n+2) to be divisible by 8. 2. n+1 is divisible by 8, when n+1 is 8,16,32,36,40,48, 56,64,72,80,88,96, but as we deal with consecutive integers, this means that in such case, n will be odd integer 7,31,35,39,47,55,63,71,79,87,95, therefore we can pick another 13 integers that make n(n+1)(n+2) divisible by eight. We have to add those 13 odds to our 47 evens. 3. n+2 is divisible by 8, when n+2 is 8,16,32,36,40,48, 56,64,72,80,88,96, but these even integers are already included in our conclusion that every even integer n from 2 to 96 satisfy the condition that n(n+1)(n+2) to be divisible by 8. we have 47+13 integers satisfying n(n+1)(n+2) is divisible by 8, thus the probability of picking one of these integers is simply P(E) = n of event occuring/total number, that is 60/96, which is equal to 5/8.



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Re: integers and prob [#permalink]
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24 Aug 2008, 14:15
marcodonzelli 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 good question. when N is even.. n(n+2) is divisable by 8 when n is odd.. n+1 must be divisible by 8 to make n(n + 1)(n + 2) probability = (96/2 + 96/8 )/96 = 60/96= 5/8
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Re: Divisibility by 8 [#permalink]
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31 Mar 2011, 01:25
The OA is wrong here, it should be D.
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Re: Divisibility by 8 [#permalink]
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31 Mar 2011, 01:31
subhashghosh wrote: The OA is wrong here, it should be D. Thanks. Corrected.
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Re: Divisibility by 8
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