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divisibility between 1 to 96

Expert replies
by r_walid » Sun Nov 01, 2009 3:20 pm
if an integer n is to be chosen at random fro mthe integers 1 to 96 inclusive, what is the probability that n(n+1)(n+2) will be divisible by 8?

1/4
3/8
1/2
5/8
3/4

please explain your reasoning.
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Source: — Problem Solving |

by sunnyjohn » Sun Nov 01, 2009 7:29 pm
IMO:D

total value of N = 96
of we can have maximum number of cases = 96

in order to divide 8, we can consider two cases:-

n = multiple of 4 + one more Even number.
n = multiple of 8

multiple of 4
4 = 2.3.4, 3.4.5, 4.5.6 = valid ==> 2.3.4, 4.5.6
8 = 6.7.8, 7.8.9, 8.9.10 <== all valid
12 = 10.11.12, 11.12.13, 12.13.14 = valid ==> 10.11.12, 12.13.14

16 = 14.15.16,15.16.17, 16.17.18
....
96 = similar cases

so we have
2 cases,
3 cases
2 cases
and so on

total case = 60

probability = 60/96 = 5/8
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by jzdchou » Mon Nov 02, 2009 10:37 am
I have 1/4

My reasoning

The 3 consecutive integer has to be even because it is either Odd*even*Odd or Even*odd*even so it is 100% divisible by 2

Then I list
1*2*3 not div by 4
2*3*4 div by 4
3*4*5 div by 4
4*5*6 div by 4
5*6*7 not div by 4
6*7*8 div by 4
7*8*9 div by 4
8*9*10 div by 4
9*10*11 not div by 4

I recognize it is NYYYNYYY basically every one out of 4 is not divisible by 4. It is 3/4 chance divisible by 4 and 100% divisibe by 2, therefore chance to be divisible by 8 = 1* 3/4 = 3/4

ANS E
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Re: divisibility between 1 to 96

by mridul_dave » Mon Nov 02, 2009 12:10 pm
r_walid wrote:if an integer n is to be chosen at random fro mthe integers 1 to 96 inclusive, what is the probability that n(n+1)(n+2) will be divisible by 8?

1/4
3/8
1/2
5/8
3/4

please explain your reasoning.
I think it is 1/2.


If n is an even integer, one of n and n+2 is divisible by 4. ( one of two consecutive even integers.
Therefore n**n+2) must be divisible by 8.
So as long as we we pick an even number from 1 through 96, n(n+1)(n+2) is divisible by 8

Since 1-96 inclusive has equal number of odd and even numbers, probability of picking an even number is 1/2.

What is the OA ?
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by Stuart@KaplanGMAT » Mon Nov 02, 2009 3:00 pm
Pattern analysis is a great way to attack this type of question - let's start there.

If we start with an even number (let's call these "even strings"), then our string of 3 integers will include both a multiple of 2 and a mutiple of 4. Therefore, every even string will be divisible by 8. That's already 50% of the strings (since half of them are even), so the answer will be at least 1/2... eliminate a, b and probably c.

If we start with an odd number (let's call these "odd strings"), then only the even number in the string could possibly be a multiple of 2. So, that number will have to account for all three 2s (since 8=2*2*2) that we need. Accordingly, if the even number in the middle of the string is a multiple of 8, an odd string will be a multiple of 8; if the even number in the middle is not a multiple of 8, then an odd string will not be a multiple of 8.

96/8 = 12, so there are 12 multiples of 8 in the bigger set. Each of those 12 numbers will appear in the middle of exactly 1 odd string, so there are 12 odd strings that are multiples of 8.

Probability = (# of desired outcomes)/(total # of possibilities)

We have 48 even strings and 12 happy odd strings, for a total of 60 strings that are multiples of 8.

We have a total of 96 strings.

Accordingly, our final answer is 60/96 = 5/8.. choose D.
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by Stuart@KaplanGMAT » Mon Nov 02, 2009 3:11 pm
We could also do this through picking numbers. Let's look at the first 8 strings:

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

Out of these 8 strings, we have 5 "yes"s and 3 "no"s; a proportion of 5/8 - pick D.

Now, you may reasonably ask "how do we know that our sample should be exactly 8 strings long?"

For questions involving multiples, the answers will almost always cycle according to what we're multiplying by. Since this question is about multiples of 8, 8 is the perfect sample size.
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Good one

by mridul_dave » Mon Nov 02, 2009 6:06 pm
Stuart Kovinsky wrote:We could also do this through picking numbers. Let's look at the first 8 strings:

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

Out of these 8 strings, we have 5 "yes"s and 3 "no"s; a proportion of 5/8 - pick D.

Now, you may reasonably ask "how do we know that our sample should be exactly 8 strings long?"

For questions involving multiples, the answers will almost always cycle according to what we're multiplying by. Since this question is about multiples of 8, 8 is the perfect sample size.

Stuart,

thanks for correcting me. I would then aproach it this way.
Either n should be even OR (n+1) should be divisible by 8

48 ways , n Can be even, and 12 ways (n+!) can be divisble by 8 ( 96/12)

48 + 12 = 60
60/96 = 5/8
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