SOLUTION: If n is an integer between 1 and 96 (inclusive), what is the probability that n(n+1)(n+2) is divisible by 8? a.1/4 b.1/2 c.5/8 d.3/4 e.7/8

Question 1204981: If n is an integer between 1 and 96 (inclusive), what is the probability that n(n+1)(n+2) is divisible by 8?
a.1/4 b.1/2 c.5/8 d.3/4 e.7/8
Found 4 solutions by ikleyn, mccravyedwin, math_tutor2020, greenestamps:
Answer by ikleyn(52835)   (Show Source): You can put this solution on YOUR website!
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If n is an integer between 1 and 96 (inclusive), what is the probability that n(n+1)(n+2) is divisible by 8?
a.1/4 b.1/2 c.5/8 d.3/4 e.7/8
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Consider the REDUCED set of numbers {1,2,3,4,5,6,7,8}. Count, how many times n*(n+1)*(n+2) is divisible by 8 for n = 1,2,3,4,5,6,7,8 T A B L E n n*(n+1)*(n+2) divisibility by 8 ------------------------------------- 1 6 2 24 (*) 3 60 4 120 (*) 5 210 6 336 (*) 7 504 (*) 8 720 (*) In the table, I marked all these cases (*) and got 5 cases of divisibility by 8. This picture will repeat cyclically every 8 consecutive positive integer numbers. So, the probability under the problem's question is 5/8. ANSWER

Solved.

Answer by mccravyedwin(408)   (Show Source): You can put this solution on YOUR website!

The triples of consecutive integers allowable are (1,2,3) through (96,97,98). So there are 96 of them Of the integers from 1 to 98, 49 are even and 49 are odd. There are only 2 types of triples of 3 consecutive integers, the (even,odd,even)-type and the (odd,even,odd)-type. Case 1. (even,odd,even)-type. Since every other even integer is divisible by 4, their product will always be divisible by 8. These are the triples (1,2,3), (3,4,5), ..., (95,96,97). There are 48 of these. [That's because the middle numbers 2,4,...,96 are such that if you divide all of them by 2 you get 1,2,...,48.] or Case 2. (odd,even,odd)-type. These will have a product divisible by 8 if and only if the middle number is divisible by 8. These are the triples (7,8,9), (15,16,17),..., (95,96,97). There are 12 of these. [That's because the middle numbers 8,16,...,96 are such that if you divide all of them by 8, you get 1,2,...,12.] So there are 48+12 = 60 triples of consecutive integers whose product is divisible by 8. The desired probability is 60/96 which reduces to 5/8 Edwin

Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

Answer: 5/8 (choice C)

Explanation

Even though the other tutors offer much more efficient solutions, I'll post mine anyway.
It follows the same idea as tutor ikleyn, but I show a bit more detail.

When dividing by 8, there are 8 possible remainders: 0 through 7.
"remainder 0" is the same as "remainder 8".

The class of numbers that give remainder 0 are: 8, 16, 24, 32, ...
i.e. the multiples of 8
Let k be an integer
n = 8k = some multiple of 8
It's fairly obvious that n(n+1)(n+2) is a multiple of 8 when n = 8k
If you aren't convinced, then replace n with 8k to see that:
n(n+1)(n+2) = 8k(8k+1)(8k+2) = 8*( k(8k+1)(8k+2) ) = 8*(some integer)

Now onto the "remainder 1" class of numbers (1, 9, 17, 25, ...)
Those are of the form n = 8k+1. They are 1 more than a multiple of 8.
Plug that into n(n+1)(n+2) to get (8k+1)(8k+2)(8k+3)
Expand that out to get the cubic 512k^3 + 384k^2 + 88k + 6
Then notice how the first three terms we can factor out 8, so we could rewrite that as
8(64k^3 + 48k^2 + 11k) + 6 = 8(some integer) + 6
Values of n in the "remainder 1" class lead to n(n+1)(n+2) having remainder 6.

Example:
n = 9
n(n+1)(n+2) = 9*10*11 = 990
990/8 = 123 remainder 6
Another example:
n = 17
n(n+1)(n+2) = 17*18*19 = 5814
5814/8 = 726 remainder 6

So we can rule out all of the "remainder 1" class numbers.

For the "remainder 2" class, we could follow similar steps to remainder 1.
But this time I'll use modular arithmetic as a shortcut.
If you aren't familiar with this notation, then please review online resources.
You could get away with doing this problem without resorting to modular arithmetic (by using the polynomial method shown above), but the polynomial method is cumbersome.

Here's what the remainder 2 case looks like
n = 8k+2
n(n+1)(n+2)
= (8k+2)(8k+2+1)(8k+2+2)
= (8k+2)(8k+3)(8k+4)
= 2*3*4 (mod 8)
= 24 (mod 8)
= 0 (mod 8)
We get remainder 0 at the end, proving that n = 8k+2 leads to n(n+1)(n+2) to be a multiple of 8.
Therefore numbers like n = 2, n = 10, n = 18, n = 26, etc will make n(n+1)(n+2) to be a multiple of 8.

Example:
n = 10
n(n+1)(n+2) = 10*11*12 = 1320
1320/8 = 165 remainder 0 = 165
1320 is a multiple of 8, so n(n+1)(n+2) is a multiple of 8 when n = 10.

I'll go through the other remainder classes fairly quickly using the mod notation
n = 8k+3
n(n+1)(n+2)
= (8k+3)(8k+3+1)(8k+3+2)
= (8k+3)(8k+4)(8k+5)
= 3*4*5 (mod 8)
= 60 (mod 8)
= 4 (mod 8)
We don't get remainder 0, so we rule out the "remainder 3" class of numbers (eg: 3, 11, 19, ...)

n = 8k+4
n(n+1)(n+2)
= (8k+4)(8k+4+1)(8k+4+2)
= (8k+4)(8k+5)(8k+6)
= 4*5*6 (mod 8)
= 120 (mod 8)
= 8*15 (mod 8)
= 0 (mod 8)
We do get remainder 0, which means any value of n part of the "remainder 4" group will make n(n+1)(n+2) a multiple of 8.
Some example values are: n = 4, n = 12, n = 20, n = 28, ...

n = 8k+5
n(n+1)(n+2)
= (8k+5)(8k+5+1)(8k+5+2)
= (8k+5)(8k+6)(8k+7)
= 5*6*7 (mod 8)
= 5*(-2)*(-1) (mod 8)
= 10 (mod 8)
= 2 (mod 8)
The nonzero remainder rules out everything in the "remainder 5" group.
Some values in this group are 5, 13, 21, ...

Cases n = 8k+6 and n = 8k+7 are handled by n = 8k-2 and n = 8k-1 respectively
But notice how if n = 8k-2 then the n+2 is a multiple of 8
Similarly, n = 8k-1 leads to n+1 being a multiple of 8
Therefore, cases n = 8k+6 and n = 8k+7 are guaranteed to lead n(n+1)(n+2) as a multiple of 8.

------------------

There's a lot to take in.
But the key summary is that we get n(n+1)(n+2) as a multiple of 8 for these remainder classes: 2, 4, 6, 7, 8
There are 5 remainder classes of values that will give us a multiple of 8.
This is out of 8 remainder classes.

Ultimately this leads to the final answer 5/8

Answer by greenestamps(13203)   (Show Source): You can put this solution on YOUR website!

If n is even, then n+2 is even, and one of them (exactly one of them) is a multiple of 4. So together the first and last of the three consecutive integers have 3 factors of 2, which makes the product divisible by 8.

Half of the numbers 1 through 96 are even, so the probability is 1/2 that n is even and the product is divisible by 8.

If n is odd, then n+2 is odd; for the product of the three consecutive integers to be divisible by 8, the middle number has to be divisible by 8. Since 1 out of every 8 integers from 1 to 96 is divisible by 8, the probability is 1/8 that n is odd and the product is divisible by 8.

The two cases of n even and n odd cover all possibilities, so the probability that the product is divisible by 8 is

ANSWER: 1/2 + 1/8 = 5/8

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