by mathslug » Wed Jan 09, 2008 4:36 am
I've simplified this problem somewhat, but I'm not sure if I've gone as far as one should go on an AIME problem.
At a glance, the problem wants us to sum the first 28 terms and take the remainder by 1000. That's cool and all, but its kinda tedious. Let's see if we can cut down on the arithmetic with some logic. Let's make use of the recursive rule.
Instead of adding a1 + a2 + a3 + a4 + a5 + a6 + a7 + a8 ... + a28, lets do:
SUM = 2(a4+a8+a12+a16+a20+a24+a28)
since a1+a2+a3=a4, a5+a6+a7=a8, etc...
To make this problem even easier, we need relationships between the terms. Lets take a look at terms a28, a29, and a30
a28 = a27 + a26 + a25
a29 = a28 + a27 + a26
a30 = a29 + a28 + a27
from this, we can derive
a30 = 2(a25) + 3(a26) + 4(a27)
Now let's use this formula we've just derived and apply it to a29, so that we can get a relationship between a28 and a24.
a29 = 2(a24) + 3(a25) + 4(a26) --> substitute a28 + a27 + a26 for a29, and we get:
a28 = 2(a24) + 3(a25) + 3(a26) -a27 --> add 4(a27) to both sides and substitute in a28:
a28 + 4(a27) = 2(a24) + 3(a28) --> The time saving eq! --> a24 + a28 = 2(a27)
This can be applied our earlier eq, and we get that the sum = 2(a4+2(a11)+2(a19)+2(a27))
I've searched for easier ways to get at a11 and a19, but the only way I could find was following the formula until I got there... : /
So the sum = 2(3+2*193+2*281+2*233) = 1417*2 = 2834 --> 834.
The answer should be 834.
Zack