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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.

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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I am still on all gmat forums. msg me if you want to ask me smth

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.

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

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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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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.

(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.

(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.
_________________

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.

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,

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

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.

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

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!!

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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30 Jun 2015, 01:11

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

The answer is D 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.

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.

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.

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