Terry has invented a new way to extend lists of numbers. To 'terryfy' a list such as [1,8] he creates 2 lists [2,9] and [3,10], where each term is one more than the corresponding term in the previous list, and then joins the three lists together to give [1,8,2,9,3,10]. If he starts with a list containing one number [0' and repeatedly 'Terryfies' it he creates the list
[0,1,2,1,2,3,2,3,4,1,2,3,2,3,4,3,4,5,2,3,4...]
What is the 2012th number in this 'Terryfy' list?
To find the 2012th number in the 'Terryfy' list, we need to analyze the pattern.
Let's observe the first few terms:
[0, 1, 2, 1, 2, 3, 2, 3, 4, 1, 2, 3, 2, 3, 4, 3, 4, 5, 2, 3, 4, ...]
We can see that the pattern starts with [0, 1, 2], and then repeats the following sub-pattern:
[1, 2, 3, 2, 3, 4, 3, 4, 5]
The length of this sub-pattern is 9. It consists of the numbers starting from 1, incrementing until a certain value, and then restarting from 2 and incrementing again. This process repeats every 9 numbers.
To find the 2012th number, we can divide 2012 by 9 and find the remainder. The quotient will give us the number of complete repetitions of the sub-pattern, and the remainder will indicate the position within the last repetition.
Dividing 2012 by 9 gives us a quotient of 224 and a remainder of 4.
Since the sub-pattern repeats every 9 numbers, the 224 repetitions would contribute a total of 224 * 9 = 2016 numbers to the 'Terryfy' list.
Therefore, the 2012th number is located within the 225th repetition of the sub-pattern and corresponds to the 4th position within that repetition.
Looking at the sub-pattern [1, 2, 3, 2, 3, 4, 3, 4, 5], the 4th position is 2.
So, the 2012th number in the 'Terryfy' list is 2.