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Similarly for \(16 = 2^4\) ( Thus total number of divisors of 8 will be \((4+1)=5\) ) which are

1, 2, 4, 8, 16. From this also we get 4 different possibilities;

\(2*(2+1)*(2+2)\) \(4*(4+1)*(4+2)\) \(8*(8+1)*(8+2)\) Two more associated with it as shown above. \(16(16+1)*(16+2)\) Two more associated with it depending on the position of 16.

Two of them are already in the sequence, so 3 different.

and so on... with each multiple of 8, we will get 3 different possibilities.

Now, how many are the total number of number which are multiple of 8 from 8...96. \((96-8/8)+1=12\)

Each of these 12 different multiples have 3 different possibilities in which 8 can be the divisor.

Thus total number of favorable outcomes = \(12*3 + 2*(2+1)*(2+2) + 4*(4+1)*(4+2) = 36+1+1\)

C 1/2 is my answer. A no can be divided by eight if we can divide that no by both 2 and 4. n(n+1)(n+2) , three consecutive nos can always be divided by 2, and if n is even, we can divide by 4 also Therefore for all even n we can divide n(n+1)(n+2) by 8 we have 48 even and 96 total nos so, prob = 48/96= 1/2 whats the OA pls?????

C 1/2 is my answer. A no can be divided by eight if we can divide that no by both 2 and 4. n(n+1)(n+2) , three consecutive nos can always be divided by 2, and if n is even, we can divide by 4 also Therefore for all even n we can divide n(n+1)(n+2) by 8 we have 48 even and 96 total nos so, prob = 48/96= 1/2

read above.

plus integers like (7,8,9) or (15,16,17)....so 1/2+1/8=5/8
_________________

C 1/2 is my answer. A no can be divided by eight if we can divide that no by both 2 and 4. n(n+1)(n+2) , three consecutive nos can always be divided by 2, and if n is even, we can divide by 4 also Therefore for all even n we can divide n(n+1)(n+2) by 8 we have 48 even and 96 total nos so, prob = 48/96= 1/2

read above.

plus integers like (7,8,9) or (15,16,17)....so 1/2+1/8=5/8

walker can you help me with this approach?

96=2^5 * 3=2^3 * 2^2 *3= so we have a total of 12 factors. to find multiples of 8 we must find all multiples of 2^3, so 6 factors, 1/2. what do I miss?

96=2^5 * 3=2^3 * 2^2 *3= so we have a total of 12 factors. to find multiples of 8 we must find all multiples of 2^3, so 6 factors, 1/2. what do I miss?

I think the approach is not correct. You missed the fact that we have 3 consecutive integers. For example, 4,5,6 are not multiplies of 8 but 4*5*6 is divisible by 8.
_________________

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

Total numbers 8*12 There are 12 numbers divisible by 8 -> 3*12 (if 8 is an example - (6,7,8), (7,8,9), (8,9,10)) and 12 numbers divisible by 4 but not divisible by 8 -> 2*12 (if 4 is an example (2,3,4) and (4,5,6))

if n is even, then n*n+2 is divisible by 8. so we have boom 50% so far.

then, we consider if n+1 is divislbe by 8, which is another 12 times

so 48 + 12 / 96 = 60/96 = 30/48 = 5/8

QED

that's a great job really. quick and easy. kudo for you

I didn't imagine it could be so easy, without any use of calculus and without trying with numbers, somewhat very difficult to do in 2 mins during the exam:

OK, here is a long explanation for a problem that took me less than a minute.

First and most important, we know that for n(n+1)(n+2) to be divisible by 8, either n or n+1 or n+2 must be divisible by 8

Second we have to realize that we are dealing with consecutive integers.

We also realize that as we have consecutive integers and multiplication, every even integer n from 2 to 96 satisfy n(n+1)(n+2) to be divisible by 8, allowing us to quickly count all even integers from 2 to 96 to be 47, when we look when n or n+1 or n+2 is divisible by 8.

Realizing the above three in a snap, we simply check when n or n+1 or n+2 is divisible by 8:

1. n is divisible by 8 when n is 8,16,32,36,40,48, 56,64,72,80,88,96, but these even integers are already included in our conclusion that every even integer n from 2 to 96 satisfy the condition that n(n+1)(n+2) to be divisible by 8.

2. n+1 is divisible by 8, when n+1 is 8,16,32,36,40,48, 56,64,72,80,88,96, but as we deal with consecutive integers, this means that in such case, n will be odd integer 7,31,35,39,47,55,63,71,79,87,95, therefore we can pick another 13 integers that make n(n+1)(n+2) divisible by eight. We have to add those 13 odds to our 47 evens.

3. n+2 is divisible by 8, when n+2 is 8,16,32,36,40,48, 56,64,72,80,88,96, but these even integers are already included in our conclusion that every even integer n from 2 to 96 satisfy the condition that n(n+1)(n+2) to be divisible by 8.

we have 47+13 integers satisfying n(n+1)(n+2) is divisible by 8, thus the probability of picking one of these integers is simply P(E) = n of event occuring/total number, that is 60/96, which is equal to 5/8.