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# If n is an integer from 1 to 96 (inclusive), what is the probability

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If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

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30 Mar 2011, 01:09
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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%

[Reveal] Spoiler:
N=n*(n+1)*(n+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!!
[Reveal] Spoiler: OA
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Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

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30 Mar 2011, 01:37
1
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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

$$Probability = \frac{48+12}{96} = \frac{60}{96}=0.625=62.5%$$

Ans: "C"

http://gmatclub.com/forum/tough-tricky-set-of-problms-85211-40.html#p799275
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Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

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30 Mar 2011, 10:43
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Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

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30 Mar 2011, 18:53
@fluke

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;

3 * 4 * 5 is not divisible by 8, right, perhaps it's more accurate to say the odd number should be at least 7 ?

Also, just to find the number of integers "(n+1)" divisible by 8 from 1 - 96 in this case :

95 = 7 + (N-1)*8 (N is total such numbers, and it's an AP)

=> (N-1)*8 = 88

=> N = 12
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Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

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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%

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Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

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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

Probability = 48+12 / 96 = 60 / 96 = 62.5%
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Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

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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 ?

Regards - Manish
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Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

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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)

n(1) = 1*2*3
n(2) = 2*3*4
n(3) = 3*4*5
n(4) = 4*5*6
n(5) = 5*6*7
n(6) = 6*7*8
n(7) = 7*8*9
n(8) = 8*9*10
n(9) = 9*10*11
n(10) = 10*11*12

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
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Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

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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)

n(1) = 1*2*3
n(2) = 2*3*4
n(3) = 3*4*5
n(4) = 4*5*6
n(5) = 5*6*7
n(6) = 6*7*8
n(7) = 7*8*9
n(8) = 8*9*10
n(9) = 9*10*11
n(10) = 10*11*12

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
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Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

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03 Apr 2011, 13:37
fluke wrote:
The product of the three consecutive number will be divisible by 8 if the first of the 3 consecutive numbers is even.

Is this something that you derived, or is this a well-known property of 8?
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Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

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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.
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Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

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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!
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Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

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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.

Here's the link where you can discuss this new question:
http://gmatclub.com/forum/hard-question-function-and-factors-90863.html
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Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

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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

$$Probability = \frac{48+12}{96} = \frac{60}{96}=0.625=62.5%$$

Ans: "C"

http://gmatclub.com/forum/tough-tricky-set-of-problms-85211-40.html#p799275

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)

and hence the required probability = 60/96=.625

please correct me if i am wrong
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Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

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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)

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.
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Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

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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)

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
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Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

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21 Apr 2011, 18:00
when n is even n (n+1)(n+2) is divisible by 8. there are 48 possible combinations for this.

n is odd and n+1 divisible by 8( i.e n = 8P-1) there are 12 possibilities that match this.

= 60/96 = 62.5

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Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

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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

so total is 5/8 = 62.5 %
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Re: If n is an integer from 1 to 96 (inclusive), what is the probability [#permalink]

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08 Oct 2015, 06:31
chloeholding wrote:
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%

[Reveal] Spoiler:
N=n*(n+1)*(n+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!!

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-an-integer-n-is-to-be-chosen-at-random-from-the-integers-126654.html
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Re: If n is an integer from 1 to 96 (inclusive), what is the probability   [#permalink] 08 Oct 2015, 06:31
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