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yes this is another neat way of looking at the problem..and you are absolutely correct that range will be divisible by 8 every 8 terms.. however, i dont think 5/8 is the exact right answer..it should be a little below 5/8

casue numbers > 8 yes every 5 numbers out of 8 will be divisible, however for number from 1-8 only 2 numbers exist 1.2.3 and 5.6.7 which are not divisble by 8..

yellowjacket wrote:

Answer is D

Suppose f(n) = n(n+1)(n+2)

We just need to calculate probability for a range of 8 numbers between 1 and 96 since every 8th number is divisible by 8 (8,16,24....)

Suppose we choose a range - 9, 10, 11, 12, 13, 14, 15, 16

Straight off, we can see that 16 is divisible. Also f(14) and f(15) are divisible becase (n+2) and (n+1) = 16 respectively.

Now f(9) and f(11) and f(13) are not divisible since they will comprise of 2 odd numbers and one even.

However, f(10) and f(12) are divisible because:

f(10) = 10*11*12 or 2*5*11*3*4 OR 8*3*5*11 Similarly for f(12).

Thus, out of 8 cases, 5 satisfy the condition of divisibility by 8.

1. If n is disvisible by 8, then n could be 8,16,24....96 ( 12 integers) 2. If n+1 is divisible by 8, then n could be 7,15,23,...95 ( 12 integers) 3. similarly for n+2

answer is D n(n+1)(n+2) will be divisible by 8 for all even numbers, i.e. total 48 numbers also cases in which (n+1) is a multiple of 8 will be divisible by 8 for ex: n=7,15,23.. i.e. total 12 numbers

so total number of such cases is 48+12=60 probability = 60/96 = 5/8

We want n(n+1)(n+2) to be divisible by 8, and that n has to be between 1 to 96. (inclusive) So for any even from 2 to 96 when substituted in the expression is divisible by 8. There are 48 even numbers till 96, inclusive. So, the probablity is (48/96) = 1/2

We want n(n+1)(n+2) to be divisible by 8, and that n has to be between 1 to 96. (inclusive) So for any even from 2 to 96 when substituted in the expression is divisible by 8. There are 48 even numbers till 96, inclusive. So, the probablity is (48/96) = 1/2

Nope.

n=7, 15, 23, etc... work as well even though they are not even. You missed those

I solved it just like Prasanna did and got 1/2. But when I looked at pisces solution I realized I missed the n+1 part. I agree with 5/8 and I liked pisces way of solving. +1 to pisces.

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

Take a range of 8 no.'s to find the no.s that will be divisible by 8 within those 8 no's which can be used as a representative for a broader range of numbers:

n(n+1)(n+2)

8x9x10 divisible by 8 9x10x11 10x11x12 divisible by 8 11x12x13 12x13x14 divisible by 8 13x14x15 14x15x16 divisible by 8 15x16x17 divisible by 8

Guys, Ans is 1/2 C.. See we have to find n so that n(n+1)(n+2)..will be div by 8.. if u select any even no as n..the above expression will be divided by 8.. so ans is 1/2...

Guys, Ans is 1/2 C.. See we have to find n so that n(n+1)(n+2)..will be div by 8.. if u select any even no as n..the above expression will be divided by 8.. so ans is 1/2...

Check the solution below. Answer is D.

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:

1. When \(n\) is even: \(n=2k\) --> \(n(n+1)(n+2)=2k(2k+1)(2k+2)=4k(2k+1)(k+1)\) --> either \(k\) or \(k+1\) is even so 8 is a multiple of \(n(n+1)(n+2)\).

2. When \(n+1\) is divisible by 8. --> \(n+1=8p\) (\(p\geq{1}\)), \(n=8p-1\) --> \(8p-1\leq{96}\) --> \(p\leq{12.1}\) --> 12 such numbers.

Also note 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: (n)*(n+1)*(n+2) is divisible by 8 when n equals: 2,4,6,7,8= 5 numbers

the probability is 5/8 ------- I think this could be enough, as 96 is a multiple of 8. So you only come to solution 60/96 = (5*12) / (8*12) _________________

now: (n)*(n+1)*(n+2) is divisible by 8 when n equals: 2,4,6,7,8= 5 numbers

the probability is 5/8 ------- I think this could be enough, as 96 is a multiple of 8. So you only come to solution 60/96 = (5*12) / (8*12)

As in EACH group of 8 numbers (1-8, 9-16, 17-24, ..., 89-96) there are EXACTLY 5 numbers for which n*(n+1)*(n+2) is divisible by 8 and 96 is a multiple of 8 (so there are integer # of such groups in the range from 1 to 96 inclusive) then the probability will be 5/8 overall, so yes your approach is correct.

now: (n)*(n+1)*(n+2) is divisible by 8 when n equals: 2,4,6,7,8= 5 numbers

the probability is 5/8 ------- I think this could be enough, as 96 is a multiple of 8. So you only come to solution 60/96 = (5*12) / (8*12)

As in EACH group of 8 numbers (1-8, 9-16, 17-24, ..., 89-96) there are EXACTLY 5 numbers for which n*(n+1)*(n+2) is divisible by 8 and 96 is a multiple of 8 (so there are integer # of such groups in the range from 1 to 96 inclusive) then the probability will be 5/8 overall, so yes your approach is correct.

Guys, Ans is 1/2 C.. See we have to find n so that n(n+1)(n+2)..will be div by 8.. if u select any even no as n..the above expression will be divided by 8.. so ans is 1/2...

Check the solution below. Answer is D.

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:

1. When \(n\) is even: \(n=2k\) --> \(n(n+1)(n+2)=2k(2k+1)(2k+2)=4k(2k+1)(k+1)\) --> either \(k\) or \(k+1\) is even so 8 is a multiple of \(n(n+1)(n+2)\).

2. When \(n+1\) is divisible by 8. --> \(n+1=8p\) (\(p\geq{1}\)), \(n=8p-1\) --> \(8p-1\leq{96}\) --> \(p\leq{12.1}\) --> 12 such numbers.

Also note 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.

Total=48+12=60

Probability: \(\frac{60}{96}=\frac{5}{8}\)

Answer: D.

Bunuel, my question involves your second bullet. What is that n=8p-1 and then 8p-1 <= 96 piece? It looks like a remainder formula but have no idea how to use the remainder formula like this??

Guys, Ans is 1/2 C.. See we have to find n so that n(n+1)(n+2)..will be div by 8.. if u select any even no as n..the above expression will be divided by 8.. so ans is 1/2...

Check the solution below. Answer is D.

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:

1. When \(n\) is even: \(n=2k\) --> \(n(n+1)(n+2)=2k(2k+1)(2k+2)=4k(2k+1)(k+1)\) --> either \(k\) or \(k+1\) is even so 8 is a multiple of \(n(n+1)(n+2)\).

2. When \(n+1\) is divisible by 8. --> \(n+1=8p\) (\(p\geq{1}\)), \(n=8p-1\) --> \(8p-1\leq{96}\) --> \(p\leq{12.1}\) --> 12 such numbers.

Also note 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.

Total=48+12=60

Probability: \(\frac{60}{96}=\frac{5}{8}\)

Answer: D.

Bunuel, my question involves your second bullet. What is that n=8p-1 and then 8p-1 <= 96 piece? It looks like a remainder formula but have no idea how to use the remainder formula like this??

Not sure I understand your question completely, but n+1 is divisible by 8 means that n is 1 less than a multiple of 8. Can you please elaborate your question? _________________

Re: If an integer n is to be chosen at random from the integers [#permalink]
16 May 2014, 10:47

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