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Hi, I'm having major problem with this question below. Can anyone help? I have seen an explanation saying that N is divisible by 8 when n is even and when (n+1) is divisible by 8 but I cannot really follow why you would approach it this way. Any advice or clearer explanation much appreciated!!
Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]
30 Mar 2011, 01:37
1
This post received KUDOS
chloeholding wrote:
Hi, I'm having major problem with this question below. Can anyone help? I have seen an explanation saying that N is divisible by 8 when n is even and when (n+1) is divisible by 8 but I cannot really follow why you would approach it this way. Any advice or clearer explanation much appreciated!!
PROBABILITY OF INTEGER BEING DIVISIBLE BY 8: If n is an integer from 1 to 96 (inclusive), what is the probability for N being divisible by 8 when N=n*(n+1)*(n+2)
A. 25% B 50% C 62.5% D. 72.5% E. 75%
N=n*(n+1)*(n+2)
Sol:
Let f(n) = n*(n+1)*(n+2)
The product of the three consecutive number will be divisible by 8 if the first of the 3 consecutive numbers is even.
f(2); the function becomes 2*3*4 will be divisible by 8 because 2 is even. f(52), the function becomes 52*53*54 will be divisible by 8 because 52 is even f(61), the function becomes 61*62*63 will NOT be divisible by 8 because 61 is odd
However, there is one more case where the function will be divisible by 8 even when first number of the three consecutive number is odd; f(7): the function becomes 7*8*9; the middle term 8 is divisible by 8 and thus the entire function becomes divisible by 8. f(63): the function becomes 63*64*65; the middle term 64 is divisible by 8 and thus the entire function becomes divisible by 8.
Probability = favorable/total
How many functions are there; f(1) = 1*2*3 f(2) = 2*3*4 f(3) = 3*4*5 f(4) = 4*5*6 . . . f(95) = 95*96*97 f(96) = 96*97*98
Total = 96
To find the favorable cases; we need to find the count of all even numbers from 1 to 96 and the count of all numbers that are divisible by 8.
Count of even numbers = (96-2)/2+1=47+1=48 Total numbers divisible by 8 = (96-8)/8+1 = 12
Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]
30 Mar 2011, 22:34
When n+1 is multiple of 8 it's just 96/8= 12 cases when f(n) = product of 3 consecutive integers is div by 8 Then when n is even we have 96/2=48 cases when f(n) is div by 8 Probability = (48+12)/96= 60/96 = 62.5%
Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]
31 Mar 2011, 20:44
PROBABILITY OF INTEGER BEING DIVISIBLE BY 8: If n is an integer from 1 to 96 (inclusive), what is the probability for N being divisible by 8 when N=n*(n+1)*(n+2)
Total number of multiples of 8 between 1 to 96, inclusive = 12 sum of even numbers = [(first even+last even) / 2] - 1 = 48
Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]
03 Apr 2011, 05:34
Hello Fluke,
I understand your question. However, I have a small doubt. In this case when we are counting the even numbers, if they have a lot of multiples of 8 already included.
In that case, aren't we over counting the possibilities ?
Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]
03 Apr 2011, 06:00
pesfunk wrote:
Hello Fluke,
I understand your question. However, I have a small doubt. In this case when we are counting the even numbers, if they have a lot of multiples of 8 already included.
In that case, aren't we over counting the possibilities ?
Regards - Manish
Not Really!!!
f(n) = n*(n+1)*(n+2)
Now, the two cases I have mentioned above are mutually exclusive. f(6) = 6*7*8. Divisible by 8 because 6 is even i.e. n is even f(7) = 7*8*9. Divisible by 8 because 7 is odd and the middle term is a multiple of 8 i.e. (n+1) is multiple of 8. f(8) = 8*9*10. Divisible by 8 because 8 is even i.e. n=even
f(6) and f(8) will be counted in the even numbers case. f(7) will be counted in the case where we are finding the (n+1) or the middle number divisibility by 8.
Let's consider a smaller set. If n is an integer from 1 to 10 (inclusive), what is the probability for N being divisible by 8 when N=n*(n+1)*(n+2)
Here there are total 10 cases. n(1) not divisible by 8 because n=1(odd) and the middle number i.e. (n+1)=2 is not divisible by 8. n(2) is divisible by 8 because n=even=2 n(3) not divisible by 8 because n=3(odd) and the middle number i.e. (n+1)=4 is not divisible by 8. n(4) is divisible by 8 because n=even=4 n(5) not divisible by 8 because n=5(odd) and the middle number i.e. (n+1)=6 is not divisible by 8. n(6) is divisible by 8 because n=even=6 n(7) is divisible by 8 because n=odd=7 and (n+1)=8 is divisible by 8 n(8) is divisible by 8 because n=even=8 n(9) not divisible by 8 because n=9(odd) but the middle number i.e. (n+1)=10 is not divisible by 8. n(10) is divisible by 8 because n=even=10
How many cases are divisible by 8: n(2) - counted for the n=even case n(4) - counted for the n=even case n(6) - counted for the n=even case n(7) - counted for the n+1 divisible by 8 case n(8) - counted for the n=even case n(10) - counted for the n=even case Count=6 out of total 10 possibilities. Do you see any repetitions here? No, right.
With formula, if we just find the number of even numbers from 1 to 10 and numbers divisible by 8. We should have our result.
Even numbers from 1 to 10 = ((10-2)/2)+1 = 5 Numbers from 1 to 10 that are divisible by 8= (8-8)/8+1=0+1=1 Total=5+1=6 _________________
Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]
03 Apr 2011, 06:29
This explains pretty well!!
Many thanks Fluke for your VERY PROMPT response
fluke wrote:
pesfunk wrote:
Hello Fluke,
I understand your question. However, I have a small doubt. In this case when we are counting the even numbers, if they have a lot of multiples of 8 already included.
In that case, aren't we over counting the possibilities ?
Regards - Manish
Not Really!!!
f(n) = n*(n+1)*(n+2)
Now, the two cases I have mentioned above are mutually exclusive. f(6) = 6*7*8. Divisible by 8 because 6 is even i.e. n is even f(7) = 7*8*9. Divisible by 8 because 7 is odd and the middle term is a multiple of 8 i.e. (n+1) is multiple of 8. f(8) = 8*9*10. Divisible by 8 because 8 is even i.e. n=even
f(6) and f(8) will be counted in the even numbers case. f(7) will be counted in the case where we are finding the (n+1) or the middle number divisibility by 8.
Let's consider a smaller set. If n is an integer from 1 to 10 (inclusive), what is the probability for N being divisible by 8 when N=n*(n+1)*(n+2)
Here there are total 10 cases. n(1) not divisible by 8 because n=1(odd) and the middle number i.e. (n+1)=2 is not divisible by 8. n(2) is divisible by 8 because n=even=2 n(3) not divisible by 8 because n=3(odd) and the middle number i.e. (n+1)=4 is not divisible by 8. n(4) is divisible by 8 because n=even=4 n(5) not divisible by 8 because n=5(odd) and the middle number i.e. (n+1)=6 is not divisible by 8. n(6) is divisible by 8 because n=even=6 n(7) is divisible by 8 because n=odd=7 and (n+1)=8 is divisible by 8 n(8) is divisible by 8 because n=even=8 n(9) not divisible by 8 because n=9(odd) but the middle number i.e. (n+1)=10 is not divisible by 8. n(10) is divisible by 8 because n=even=10
How many cases are divisible by 8: n(2) - counted for the n=even case n(4) - counted for the n=even case n(6) - counted for the n=even case n(7) - counted for the n+1 divisible by 8 case n(8) - counted for the n=even case n(10) - counted for the n=even case Count=6 out of total 10 possibilities. Do you see any repetitions here? No, right.
With formula, if we just find the number of even numbers from 1 to 10 and numbers divisible by 8. We should have our result.
Even numbers from 1 to 10 = ((10-2)/2)+1 = 5 Numbers from 1 to 10 that are divisible by 8= (8-8)/8+1=0+1=1 Total=5+1=6
Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]
03 Apr 2011, 13:48
plb wrote:
fluke wrote:
The product of the three consecutive number will be divisible by 8 if the first of the 3 consecutive numbers is even. Yes, because we will have product of two consecutive even numbers included in the expression.
Is this something that you derived, or is this a well-known property of 8?
It is a well known property. The product of two consecutive even numbers will always be divisible by 8.
Here's why; 8's prime factors = 2*2*2 2's prime factors = 2 4's prime factors = 2*2 2*4 prime factors = 2*2*2
Even if we don't take specifically 2 or 4. For any two consecutive even numbers; 1 number will be divisible by 2 and another divisible by 4.
Any two consecutive even numbers will definitely have "2*2*2" as its factors. And 8 also has "2*2*2" as its factor. Thus, the product of two consecutive even integers will always be divisible by 8. _________________
Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]
04 Apr 2011, 06:08
Hi all, thanks so much for your replies, has been very helpful. I have a new problem though...
Question: For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n inclusive. If p is the smallest prime factor of h(100)+1, then p is a) between 2 and 10 b) between 10 and 20 c) between 20 and 30 d) between 30 and 40 e) greater than 40
Any replies would be greatly appreciated - thanks!
Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]
04 Apr 2011, 06:28
chloeholding wrote:
Hi all, thanks so much for your replies, has been very helpful. I have a new problem though...
Question: For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n inclusive. If p is the smallest prime factor of h(100)+1, then p is a) between 2 and 10 b) between 10 and 20 c) between 20 and 30 d) between 30 and 40 e) greater than 40
Any replies would be greatly appreciated - thanks!
chloeholding: Please post only one question per thread. The discussion for a new question should be opened in a separate thread.
Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]
05 Apr 2011, 06:43
fluke wrote:
chloeholding wrote:
Hi, I'm having major problem with this question below. Can anyone help? I have seen an explanation saying that N is divisible by 8 when n is even and when (n+1) is divisible by 8 but I cannot really follow why you would approach it this way. Any advice or clearer explanation much appreciated!!
PROBABILITY OF INTEGER BEING DIVISIBLE BY 8: If n is an integer from 1 to 96 (inclusive), what is the probability for N being divisible by 8 when N=n*(n+1)*(n+2)
A. 25% B 50% C 62.5% D. 72.5% E. 75%
N=n*(n+1)*(n+2)
Sol:
Let f(n) = n*(n+1)*(n+2)
The product of the three consecutive number will be divisible by 8 if the first of the 3 consecutive numbers is even.
f(2); the function becomes 2*3*4 will be divisible by 8 because 2 is even. f(52), the function becomes 52*53*54 will be divisible by 8 because 52 is even f(61), the function becomes 61*62*63 will NOT be divisible by 8 because 61 is odd
However, there is one more case where the function will be divisible by 8 even when first number of the three consecutive number is odd; f(7): the function becomes 7*8*9; the middle term 8 is divisible by 8 and thus the entire function becomes divisible by 8. f(63): the function becomes 63*64*65; the middle term 64 is divisible by 8 and thus the entire function becomes divisible by 8.
Probability = favorable/total
How many functions are there; f(1) = 1*2*3 f(2) = 2*3*4 f(3) = 3*4*5 f(4) = 4*5*6 . . . f(95) = 95*96*97 f(96) = 96*97*98
Total = 96
To find the favorable cases; we need to find the count of all even numbers from 1 to 96 and the count of all numbers that are divisible by 8.
Count of even numbers = (96-2)/2+1=47+1=48 Total numbers divisible by 8 = (96-8)/8+1 = 12
first i counted the no which are divisible by 8 through1 - 96 it was 96/8=12 so for each value there will be 3 pairs I.e. 12 x 3 =36 similarly i counted the nos which are divisible by 4 it was 96/4=24 now (as if n = multiple of 4 then n+2 will be even hence multiple of 8)
i added up 36+24=60 and hence the required probability = 60/96=.625
please correct me if i am wrong _________________
WarLocK _____________________________________________________________________________ The War is oNNNNNNNNNNNNN for 720+ see my Test exp here http://gmatclub.com/forum/my-test-experience-111610.html do not hesitate me giving kudos if you like my post.
Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]
05 Apr 2011, 07:35
Warlock007 wrote:
Dear fluke I did it by simple method as below
first i counted the no which are divisible by 8 through1 - 96 it was 96/8=12 so for each value there will be 3 pairs I.e. 12 x 3 =36 similarly i counted the nos which are divisible by 4 it was 96/4=24 now (as if n = multiple of 4 then n+2 will be even hence multiple of 8)
i added up 36+24=60 and hence the required probability = 60/96=.625
please correct me if i am wrong
You got the answer right but there seems some discrepancy in the counting.
When you counted all numbers divisible by 8 and then multiplied by 3; that would include sets such as
Below; 8 is divisible by 8 6*7*8 7*8*9 8*9*10
Below; 16 is divisible by 8 14*15*16 15*16*17 16*17*18
Below; 80 is divisible by 8 78*79*80 79*80*81 80*81*82
Sets where n=4 or a multiple of 4, such as 4*5*6 8*9*10 12*13*14 16*17*18
The colored sets are counted twice each.
At the same time, the following sets are never counted: 2*3*4 10*11*12
Luckily, the number of sets you missed compensated for the number of sets you double counted and got the correct result. _________________
Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]
05 Apr 2011, 07:38
fluke wrote:
Warlock007 wrote:
Dear fluke I did it by simple method as below
first i counted the no which are divisible by 8 through1 - 96 it was 96/8=12 so for each value there will be 3 pairs I.e. 12 x 3 =36 similarly i counted the nos which are divisible by 4 it was 96/4=24 now (as if n = multiple of 4 then n+2 will be even hence multiple of 8)
i added up 36+24=60 and hence the required probability = 60/96=.625
please correct me if i am wrong
You got the answer right but there seems some discrepancy in the counting.
When you counted all numbers divisible by 8 and then multiplied by 3; that would include sets such as
Below; 8 is divisible by 8 6*7*8 7*8*9 8*9*10
Below; 16 is divisible by 8 14*15*16 15*16*17 16*17*18
Below; 80 is divisible by 8 78*79*80 79*80*81 80*81*82
Sets where n=4 or a multiple of 4, such as 4*5*6 8*9*10 12*13*14 16*17*18
The colored sets are counted twice each.
At the same time, the following sets are never counted: 2*3*4 10*11*12
Luckily, the number of sets you missed compensated for the number of sets you double counted and got the correct result.
hey fluke you are right man....Thanks for your help _________________
WarLocK _____________________________________________________________________________ The War is oNNNNNNNNNNNNN for 720+ see my Test exp here http://gmatclub.com/forum/my-test-experience-111610.html do not hesitate me giving kudos if you like my post.
Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]
08 Oct 2015, 01:00
chloeholding wrote:
Hi, I'm having major problem with this question below. Can anyone help? I have seen an explanation saying that N is divisible by 8 when n is even and when (n+1) is divisible by 8 but I cannot really follow why you would approach it this way. Any advice or clearer explanation much appreciated!!
PROBABILITY OF INTEGER BEING DIVISIBLE BY 8: If n is an integer from 1 to 96 (inclusive), what is the probability for N being divisible by 8 when N=n*(n+1)*(n+2)
A. 25% B 50% C 62.5% D. 72.5% E. 75%
N=n*(n+1)*(n+2)
either n is divisible by 8 so 12 factors of 8 in the list, 12/96 = 1/8
or n is even which means 48 even numbers in list 48 /96 = 1/2
Hi, I'm having major problem with this question below. Can anyone help? I have seen an explanation saying that N is divisible by 8 when n is even and when (n+1) is divisible by 8 but I cannot really follow why you would approach it this way. Any advice or clearer explanation much appreciated!!
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