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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
_________________

Happy are those who dream dreams and are ready to pay the price to make them come true

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.

Re: If an integer n is to be chosen at random from the integers [#permalink]
18 Feb 2012, 07:03

2

This post received KUDOS

+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
_________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

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.

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

(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 [#permalink]
19 Feb 2013, 01:15

1

This post received KUDOS

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.............

Re: If an integer n is to be chosen at random [#permalink]
19 Feb 2013, 02:25

Expert's post

mun23 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+) will be divisible by8? (A)1/4 (B)3/8 (C)1/2 (D)5/8 (E)3/4 Need explanation.............

Merging similar topics. Please ask if anything remains unclear.
_________________

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]
20 Aug 2013, 07:11

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]
25 Sep 2013, 03:17

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.

gmatclubot

Re: If an integer n is to be chosen at random from the integers
[#permalink]
25 Sep 2013, 03:17