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If an integer n is to be chosen at random from the integers [#permalink]
 16 Apr 2008, 10:25
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54% (01:03) wrong based on 35 sessions
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
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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.
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i really didnt do any math on this...i realize as long as n=even its divisible by 8..
so right away i know that probability is greater than 1/2
now..3.4.5 is divisible too..soo really its down to D and E..
then i notice 9.10.11 is not divisible..
basically how many n or n+1 or n+2=2^3*M
5/8 makes sense..
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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.
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i think ans is 3/8.
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
There probability = (12+12+12)/96
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Since 96 is divisible by 8 and since "divisibility by 8" repeats every 8 terms, you can just focus on the 8 first terms (from 1 to 8):
it works for n=2,4,6,7,8, that is 5 numbers out of the 8
==> Answer is (D) = 5/8
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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
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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
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Prasanna1981 wrote: 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 
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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.
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JCLEONES 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 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 Probability = 5/8
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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...
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sudhanshushankerjha wrote: 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). # of even numbers between 1 and 96, inclusive is \frac{96-2}{2}+1=48 (check this: totally-basic-94862.html?hilit=last%20first%20range%20multiple) AND 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.
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Could you comment on my sollution please? Write numbers: 1,2,3,4,5,6,7,8 = 8 numbers 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)
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craky wrote: Could you comment on my sollution please?
Write numbers:
1,2,3,4,5,6,7,8 = 8 numbers
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. Check this question for similar solution: beginner-s-forum-question-106168.html
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Bunuel wrote: craky wrote: Could you comment on my sollution please?
Write numbers:
1,2,3,4,5,6,7,8 = 8 numbers
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. Check this question for similar solution: beginner-s-forum-question-106168.htmlWhoau, it looks I somehow discovered your 30 second approach. Thanks for clarification.
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Bunuel wrote: sudhanshushankerjha wrote: 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). # of even numbers between 1 and 96, inclusive is \frac{96-2}{2}+1=48 (check this: totally-basic-94862.html?hilit=last%20first%20range%20multiple) AND 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??
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jmuduke08 wrote: Bunuel wrote: sudhanshushankerjha wrote: 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). # of even numbers between 1 and 96, inclusive is \frac{96-2}{2}+1=48 (check this: totally-basic-94862.html?hilit=last%20first%20range%20multiple) AND 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 farther?
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