# Worded Problem - Difficult to solve

• Nov 9th 2012, 03:06 PM
sammukeshav
Worded Problem - Difficult to solve
Terry has invented a new way to extend lists of numbers. To Terryfy a list such as [1, 8] he creates two 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 Terryfic list?
• Nov 9th 2012, 04:02 PM
jakncoke
Re: Worded Problem - Difficult to solve
The 2012 number is 1283. Note this could be described by the function, which tells us the number at position x in the sequence, \$\displaystyle f(x) = x mod 3^n\$ where n = floor[x/3].

Think of this problem like a counter, which grows by size 3^n, the first set {0} = 3^0 = 1
The second set \$\displaystyle {0, 1, 2} = 3^1\$
The third set \$\displaystyle {0, 1, 2, 1, 2, 3, 2, 3, 4} = 3^2\$
....

As soon as the counter resets, At \$\displaystyle 3^0 , 3^1, 3^2\$, we me make it bigger \$\displaystyle 3^(n+1)\$, and start counting again from 0 till the counter resets, that is at \$\displaystyle 3^(n+1) \$