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If an integer n is to be chosen at random from the integers

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

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New post 20 Sep 2015, 12:30
take number 1,2,,3,4,5,6,7,8.

pick any one and check for n(n+1)(n+2), only 5 favourable cases. so 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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New post 08 Oct 2015, 10:49
Hi All,

For a number to be evenly divisible by 8, it has to include at least three 2's when you prime factor it.

For example,
8 is divisible by 8 because 8 = (2)(2)(2).....it has three 2s "in it"
48 is divisible by 8 because 48 = (3)(2)(2)(2)(2).....it has three 2s "in it" (and some other numbers too).

20 is NOT divisibly by 8 because 20 = (2)(2)(5)....it only has two 2s.

In this question, when you take the product of 3 CONSECUTIVE POSITIVE INTEGERS, you will either have....

(Even)(Odd)(Even)

or

(Odd)(Even)(Odd)

In the first option, you'll ALWAYS have three 2s. In the second option, you'll only have three 2s if the even term is a multiple of 8 (Brent's list proves both points). So for every 8 consecutive sets of possibilities, 4 of 4 from the first option and 1 of 4 from the second option will give us multiples of 8. That's 5/8 in total.

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

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New post 16 Jul 2016, 02:40
Bunuel wrote:
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


\(n(n + 1)(n + 2)\) 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.

Answer: D.

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

Hope it helps.


Hi,

Thank you for posting such a good explanation , however I could not understand how can you categorize the numbers in group of 8 . Probability is Fav/Total . Shouldn't we consider all 96 values and find out how many are satisfying our conditions , cant understand how you arrive at 5/8. If you can please explain.
Sorry if this sounds too basic.

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

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New post 16 Jul 2016, 02:41
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megha_2709 wrote:
Bunuel wrote:
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


\(n(n + 1)(n + 2)\) 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.

Answer: D.

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

Hope it helps.


Hi,

Thank you for posting such a good explanation , however I could not understand how can you categorize the numbers in group of 8 . Probability is Fav/Total . Shouldn't we consider all 96 values and find out how many are satisfying our conditions , cant understand how you arrive at 5/8. If you can please explain.
Sorry if this sounds too basic.

Regards
Megha


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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New post 16 Jul 2016, 06:44
Bunuel wrote:
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


\(n(n + 1)(n + 2)\) 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.

Answer: D.

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

Hope it helps.

i have a doubt , please help me what m i missing here.
when you are considering pair 1-8 actually you are grouping
(2,3,4),(4,5,6),(6,7,8),(8,9,10) and 7,8,9
we are takig 8 first position because ...in second group first pair will be 10,11,12
similarly to make the count 5 for each pair , we have to consider 96 at first position 96,97,98
Is it not increasing out given number limit.
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Re: If an integer n is to be chosen at random from the integers [#permalink]

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New post 24 Jul 2016, 05:02
Bunuel

Thanks for replying and clearing m doubt. Really helped. :)

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

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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


We are given that an integer n is to be chosen at random from the integers 1 to 96 inclusive, and we need to determine the probability that n(n + 1)(n + 2) will be divisible by 8.

We should recall that when a number is divisible by 8, it is divisible by 2^3, i.e., three factors of 2. We should also recognize that n(n + 1)(n + 2) is the product of three consecutive integers.

Case 1: n is even. Any time that n is even, n + 2 will also be even. Moreover, either n or n + 2 will be divisible by 4, and thus n(n + 1)(n + 2) will contain three factors of 2 and will be divisible by 8.

Since there are 96 integers between 1 and 96, inclusive, and half of those integers are even, there are 48 even integers (i.e., 2, 4, 6, …, 96) from 1 to 96 inclusive. Thus, when n is even, there are 48 instances in which n(n + 1)(n + 2) will be divisible by 8.

Case 2: n is odd. If n is odd, then n(n + 1)(n + 2) still can be divisible by 8 if the factor (n + 1) is a multiple of 8. So, let’s determine the number of multiples of 8 between 1 and 96 inclusive.

Number of multiples of 8 = (96 - 8)/8 + 1 = 88/8 + 1 = 12. Thus, when n is odd, there are 12 instances in which n(n + 1)(n + 2) will be divisible by 8.

In total, there are 48 + 12 = 60 outcomes in which n(n + 1)(n + 2) will be divisible by 8.

Thus, the probability that n(n + 1)(n + 2) is divisible by 8 is: 60/96 = 10/16 = 5/8.

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

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New post 24 Jan 2017, 18:33
1)Probability of an event = Number of Desired Outcomes/Number of Possible Outcomes
2)A number is divisible by 8 if three of its prime factors are 2's.
3)If n is even, then n+2 is also even, and there are at least three 2's between these two multiples. For example, if n=2, then 2*(2+2)=8=2*2*2. This means that at least in 50% of cases, the product n(n+1)(n+2) is divisible by 8.
4)If n is odd, then the only way it is divisible by 8 is if (n+1) is divisible by 8. To calculate the total number of multiples of 8 between 1 and 96: (96-8)/8+1=88/8+1=12
5)96/2+12=60; 60/96=5/8

The correct answer is D.
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Re: If an integer n is to be chosen at random from the integers [#permalink]

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My approach was similar to Bunuel's and AdmitJA's, but I wanted to offer it because it was simple and quick. I was able to do this one in just over a minute.

When I read the question, I paid attention to two things right away. The first was that \((n)(n+1)(n+2)\) is the product of three consecutive integers. Because \(8=2^3\), I knew that I needed to find three 2s in the factorization of whatever 3 consecutive numbers I used. The second thing that I noticed is that 96 is divisible by 8. This indicated to me that whatever pattern I noticed in the first 8 numbers would be repeated 12 times through 96. Since I recognized that the pattern would be repeated, I knew that I only needed to look at the first 8 numbers.

I listed the numbers out: 1 2 3 4 5 6 7 8.

Since 8 has three 2s in its factorization, I knew that n=6,7,8 would all be divisible by 8. I saw that n=2 would work because 2 has one 2 and 4 has two 2s. Similarly, I saw that n=4 would work because 4 has two 2s and 6 has one 2. That gives us 5 options in the first 8 numbers. So the answer is 5/8.

As mentioned above, this ratio will be the same for every 8 numbers, so 5/8 will be true of 8x where x is any positive integer. A trickier version of this question would have been to make the number NOT divisible by 8. In that case, I think you should still find the pattern for every 8 numbers, but you'd also want to look at the "extra" numbers to figure out the fraction.

Take 53, for instance. You'd want to recognize that 8 goes into 53 six times with a remainder of 5. This means that you'd have the \(5/8\) ratio for 6 sets of 8 but also an "incomplete" set of n = 1,2,3,4,5. In the first 48 numbers, you'd have 30 that would be divisible by 8. In the "incomplete" set you'd have 2 (since n=2 and n=4 are both divisible by 8). Thus you'd have 32/53 numbers divisible by 8.
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If an integer n is to be chosen at random from the integers [#permalink]

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New post 29 Aug 2017, 19:33
Here's my take, please correct me if I were wrong.

multiply out n(n+1)(n+2) = n^3 + 3n^2 + 2n.

Now, let's factor out n^3 from above equation which become n^3 (1+ 3n^-1 + 2n^-2) <-- I know they looks ugly but wait

You basically ignore (1+ 3n^-1 + 2n^-2) now so that you will know that to have n^3 divisible by 8; n must be the multiple of 2.

Here you will know that 96 / 2 = 48 numbers that are the multiple of 2.

Be aware here because 48 number have included the multiple of 8's but we still need to take (n+1) into consideration. So total number of (n+1 = 8) are 12.

therefore (48+12) / 96 = 5/8.
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If an integer n is to be chosen at random from the integers [#permalink]

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New post 15 Oct 2017, 08:26
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

My method- 1,2,3 ->NO: 2,3,4->Yes: 3,4,5->NO; 4,5,6-> Yes: 5,6,7- NO: 6,7,8->Yes, 7,8,9->Yes: 8,9,10:> Yes

Wait a sec! what just happened? whenever there is 8(or a multiple of 9, irrespective of the other two numbers, the overall number becomes divisible by 8 and even the last number, 96, is a multiple of 8. So, 5/8
is the answer. Don't believe me? Count them!
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Re: If an integer n is to be chosen at random from the integers [#permalink]

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New post 15 Oct 2017, 10:33
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


This is a really interesting question and is of 700 level..

So, there might be two cases..
Case 1: n is even .. So if n is a multiple of 2 and not 4 (for e.g 2,6,10..) then n+2 must be multiple of 4 (respectively 4,8,12..) Hence , n(n+2) must be divisible by 8 and hence n(n+1)(n+2) must be divisible by 8.
Similarly if n is a multiple of 4 (for e.g 4,8,12..) then n+2 must be multiple of 4 (respectively 6,10,14..) Hence , n(n+2) must be divisible by 8 and hence n(n+1)(n+2) must be divisible by 8.

So, in this case n(n+1)(n+2) must be divisible by 8.

Such numbers are 96/2 = 48

Case 2: n is odd.. So, (n+1) is even and n+2 is odd.. n(n+1)(n+2) can be divisible by 8 only if (n+1) is a multiple of 8 .

So, (n+1) must be (8,16,...96) and corresponding values of n will be (7,15,..., 95)

Such numbers are 96/8 = 12

Probability that n(n + 1)(n + 2) will be divisible by 8 = (48+12)/96 = 60/96 = 5/8

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

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New post 19 Oct 2017, 22:26
Bunuel wrote:
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


\(n(n + 1)(n + 2)\) 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.

Answer: D.

Similar question: http://gmatclub.com/forum/divisible-by- ... 21561.html

Hope it helps.


Hi Bunuel
Why are we choosing a bracket of 8 numbers??? pls explain..
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Re: If an integer n is to be chosen at random from the integers [#permalink]

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New post 19 Oct 2017, 23:14
zanaik89 wrote:
Bunuel wrote:
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


\(n(n + 1)(n + 2)\) 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.

Answer: D.

Similar question: http://gmatclub.com/forum/divisible-by- ... 21561.html

Hope it helps.


Hi Bunuel
Why are we choosing a bracket of 8 numbers??? pls explain..


Because the pattern is repeated in every 8 numbers.
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Re: If an integer n is to be chosen at random from the integers [#permalink]

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New post 20 Apr 2018, 13:32
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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


First recognize that n, n+1 and n+2 are 3 CONSECUTIVE INTEGERS.

Now let's make some observations:

When n = 1, we get: (1)(2)(3), which is NOT divisible by 8
n = 2, we get: (2)(3)(4), which is DIVISIBLE BY 8
n = 3, we get: (3)(4)(5), which is NOT divisible by 8
(4)(5)(6), which is DIVISIBLE BY 8
(5)(6)(7), which is NOT divisible by 8
(6)(7)(8), which is DIVISIBLE BY 8
(7)(8)(9), which is DIVISIBLE BY 8
(8)(9)(10), which is DIVISIBLE BY 8
-----------------------------
(9)(10)(11), which is NOT divisible by 8
(10)(11)(12), which is DIVISIBLE BY 8
(11)(12)(13), which is NOT divisible by 8
(12)(13)(14), which is DIVISIBLE BY 8
(13)(14)(15), which is NOT divisible by 8
(14)(15)(16), which is DIVISIBLE BY 8
(15)(16)(17), which is DIVISIBLE BY 8
(16)(17)(18)which is DIVISIBLE BY 8
-----------------------------
.
.
.
The pattern tells us that 5 out of every 8 products is divisible by 8.
So, 5/8 of the 96 products will be divisible by 8.
This means that the probability is 5/8 that a given product will be divisible by 8.

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

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New post 25 Apr 2018, 16:43
I looked at this problem as sets.
I divided 96/8 = 16. So I have 16 sets of 8 consecutive numbers (1-8), (9-16)...(89-96). So I focused on the 1st set (1-8).
n(n+1)(n+2) are 3 consecutive numbers being multiplied. So they will always be divisible by 3. But the question asks for divisibility by 8. When you prime factorize 8, you end up with 2^3. Focusing on set 1-8, I focused on multiplying 3 consecutive numbers to see if I can pull out at minimum 2^3.

1*2*3 = No (there is a single 2)
2*3*4 = yes ( there are three 2's)
3*4*5 = no ( there are two 2's)
4*5*6 = yes
5*6*7 = No
6*7*8 = yes
7*8*9 = yes
8*9*10 = yes

You have 5 yes out of 8 in the set. 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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New post 15 May 2018, 10:06
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


VeritasPrepKarishma Pls show us your approach to this question. Though I understood the solutions mentioned in the thread, I am finding it difficult to believe that I might think of this approach during actual exam.
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Re: If an integer n is to be chosen at random from the integers [#permalink]

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New post 15 May 2018, 10:26
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siddreal wrote:
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


VeritasPrepKarishma Pls show us your approach to this question. Though I understood the solutions mentioned in the thread, I am finding it difficult to believe that I might think of this approach during actual exam.


What I would do is pretty much what Bunuel has done in his solution.

Note that the moment I see n(n + 1)(n + 2), I think of divisibility in a bunch of consecutive numbers.
In 3 consecutive integers, if n is even, (n+2) is even too. If there are two consecutive even integers, one of them will be a multiple of 4.
So if n is even, (n+2) is even too and one of them is definitely a multiple of 4.
So n(n + 1)(n + 2) becomes divisible by 8 in each case that n is even.
From 1 to 96, half the cases have even n so this mean 48 cases.

Alternatively, when n is odd, n+1 is even. But then n+2 is odd too. So to be a multiple of 8, (n+1) will need to be a multiple of 8.
Hence this gives us another 12 cases (n+1 goes from 8 to 96).
Note that there will be no overlap in the two since here n is definitely odd.

Total we have 60 cases of the possible 96 which gives 60/96 = 5/8

For these properties of numbers, see
https://www.veritasprep.com/blog/2011/0 ... c-or-math/
https://www.veritasprep.com/blog/2011/0 ... h-part-ii/
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Re: If an integer n is to be chosen at random from the integers   [#permalink] 15 May 2018, 10:26

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