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

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If an integer n is to be chosen at random from the integers [#permalink] New post 16 Apr 2008, 09:25
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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
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Re: PS Integers [#permalink] New post 16 Apr 2008, 09:45
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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Re: PS Integers [#permalink] New post 16 Apr 2008, 10:56
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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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Re: PS Integers [#permalink] New post 16 Apr 2008, 11:00
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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Re: PS Integers [#permalink] New post 09 Jul 2008, 00:12
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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Re: PS Integers [#permalink] New post 09 Jul 2008, 00:33
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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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Re: PS Integers [#permalink] New post 09 Jul 2008, 02:49
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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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Re: PS Integers [#permalink] New post 09 Jul 2008, 04:52
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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Re: PS Integers [#permalink] New post 09 Jul 2008, 04:56
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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Re: PS Integers [#permalink] New post 09 Jul 2008, 05:09
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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Re: PS Integers [#permalink] New post 22 Sep 2009, 19:39
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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Re: PS Integers [#permalink] New post 18 Oct 2010, 10:26
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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Re: PS Integers [#permalink] New post 19 Oct 2010, 11:43
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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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Re: PS Integers [#permalink] New post 02 Feb 2011, 06:23
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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Re: PS Integers [#permalink] New post 02 Feb 2011, 06:48
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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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Re: PS Integers [#permalink] New post 02 Feb 2011, 07:02
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.html


Whoau, it looks I somehow discovered your 30 second approach. Thanks for clarification.
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Re: PS Integers [#permalink] New post 26 Mar 2013, 14:08
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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Re: PS Integers [#permalink] New post 27 Mar 2013, 04:52
Expert's post
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?
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post 16 May 2014, 10:47
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Re: If an integer n is to be chosen at random from the integers   [#permalink] 16 May 2014, 10:47
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