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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]
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Re: If an integer n is to be chosen at random from the integers [#permalink]
20 Dec 2015, 04:41
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