# AMC Question HARD

1. Aug 5, 2010

### asdfghjklqqww

AMC Question HARD!!!

While doing the Year Eight AMC (Australian Maths Competition), I came across this diffucult question...
(By the way, this isn't homework :P)

"I have a list of twelve numbers. The first number is 1 and the last number is 12. Each of the numbers is one more than the average of its neighbours. What is the heighest number on the list?"

It was an open answered question, and I couldn't work it out, so with seconds to spare, I guessed the number 47.

This is most likely wrong, so what is the actual answer???

2. Aug 5, 2010

### Petr Mugver

Re: AMC Question HARD!!!

Your numbers are (I omit the calculations, but you can verify the answer is correct)

$$a_n=1+12n-n^2\qquad n=0,1,\dots,10,11$$

So

a_0=1
a_1=12
a_2=21
a_3=28
a_4=33
a_5=36
a_6=37
a_7=36
a_8=33
a_9=28
a_10=21
a_11=12

You was wrong, largest number is 37.

3. Aug 18, 2010

### ross_tang

Re: AMC Question HARD!!!

You may just write out the recurrence equation and solve it yourself:

$$a_n = \frac{a_{n+1}+a_{n-1}}{2} +1$$

$$a_1=1$$

$$a_{12}=12$$

The solution is:

$$a_n = -n^2 + 14 n -12$$