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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 [#permalink]

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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 28 Jan 2012, 05:12, edited 1 time in total.

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28 Jan 2012, 05:12
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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

$$n(n + 1)(n + 2)$$ 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.

Similar question: divisible-by-12-probability-121561.html

Hope it helps.
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29 Jan 2012, 01:17
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Bunuel 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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30 Jan 2012, 09:10
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Bunuel 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.

Similar question: divisible-by-12-probability-121561.html

Hope 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]

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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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06 May 2012, 02:22
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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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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

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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08 May 2012, 01:05
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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

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 [#permalink]

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17 May 2013, 13:07
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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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19 Aug 2013, 01:39
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Bunuel 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.

Similar question: divisible-by-12-probability-121561.html

Hope 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 [#permalink]

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20 Aug 2013, 08:11
We can see that all even values of n will be divisible by 8 as one will be multiple of 2 and other of 4

So, even numbers b/w 1 and 96 inclusive=48
Secondly we can see that any n+1 which is a multiple of 8 also satisfies this equation, so the value of n can be 7,15, ... ,95 therefore numbers in this range 95-7/8+1=12

Hence,total fav outcomes=48+12=60
therefore prob=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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25 Sep 2013, 04:17
Hi! Can this question be solved using the LCM approach? We need n(n+1)(n+2) to be divisible by 8. The number will also be divisible by 3. So is there any way we can do it using the LCM of 3 and 8?
Or should we just leave out 3 because for all vales of n, n(n+1)(n+2) will be divisible by 3.

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If an integer n is to be chosen at random from the integers [#permalink]

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Re: If an integer n is to be chosen at random from the integers [#permalink]

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17 Aug 2014, 00:24
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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 1-98, 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 [#permalink]

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04 Sep 2014, 13:42
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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.

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Re: If an integer n is to be chosen at random from the integers [#permalink]

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05 Sep 2014, 06:05
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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 [#permalink]

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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:

Capture.PNG [ 6.47 KiB | Viewed 44096 times ]

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Re: If an integer n is to be chosen at random from the integers [#permalink]

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30 Jun 2015, 01:11
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

There are two ways of solving this problem.
1) the product of three consecutive numbers must be divisible by 8. when n is odd n+2 is also odd. Therefore, n+1 has to be divisible by 8. If you check the first 8 set of numbers, 5 are divisible by 8. In addition 96 is also divisible by 8. Hence, the answer is 5/8.
2)alternatively, n is odd in 48 cases. Out of this n+1 is divisible by 8 in 12 cases. Therefore, 36 cases are not divisible by 8. when n is even all the 48 cases are divisible by 8. Hence, 60/96 or 5/8 is the required answer.

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Re: If an integer n is to be chosen at random from the integers [#permalink]

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12 Jul 2015, 05:40
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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

Alternate solution:

Look at the first few sets of 3s:
1,2,3
2,3,4
3,4,5
4,5,6
5,6,7
6,7,8
7,8,9
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
90,91,92
91,92,93
92,93,94
93,94,95
94,95,96
95,96,97
96,97,98

So 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: If an integer n is to be chosen at random from the integers [#permalink]

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21 Aug 2015, 02:44
the official explanation seems to be too ambigious - i like the explations above given by Bunuel and metall
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, (48+12)96=58

D
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Re: If an integer n is to be chosen at random from the integers   [#permalink] 21 Aug 2015, 02:44

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