Question 1204981
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Answer:   <font color=red size=4>5/8  (choice C)</font>


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.


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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 <font color=red size=4>final answer 5/8</font>
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