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integers and prob

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integers and prob [#permalink] New post 19 Jan 2008, 05:19
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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] New post 19 Jan 2008, 12:18
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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 8n-2, 8n-1, 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] New post 19 Jan 2008, 12:33
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 8n-2, 8n-1, 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 1-96 when n(n+1)(n+2) is divisible by 8.
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Re: integers and prob [#permalink] New post 19 Jan 2008, 12:41
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ah... you are right :wall

D

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 8n-2, 8n-1, 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 8n-6, 8n-4

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] New post 19 Jan 2008, 12:43
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GMATBLACKBELT wrote:
I get every even number from 1-96 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] New post 19 Jan 2008, 22:01
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.
(96-8/8)+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] New post 24 Jan 2008, 10:43
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] New post 24 Jan 2008, 10:49
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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] New post 20 Feb 2008, 00:46
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] New post 20 Feb 2008, 01:35
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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] New post 20 Feb 2008, 09:20
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] New post 20 Feb 2008, 11:24
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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] New post 20 Feb 2008, 22:56
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] New post 17 Mar 2008, 02:32
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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] New post 28 Apr 2008, 00:09
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. :wink:
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Re: integers and prob [#permalink] New post 24 Aug 2008, 13:15
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] New post 31 Mar 2011, 00:25
The OA is wrong here, it should be D.
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Re: Divisibility by 8 [#permalink] New post 31 Mar 2011, 00:31
subhashghosh wrote:
The OA is wrong here, it should be D.


Thanks. Corrected.
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Re: Divisibility by 8   [#permalink] 31 Mar 2011, 00:31
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