IBM Research | Ponder This | January 2012 solutions

# Ponder This

## January 2012

<<December January February>>

In the first part of the challenge, the dwarfs leave the scene after a second and leave a lamp on if and only if its number has an odd number of divisors.

Divisors usually comes in pairs: d and n/d. Only perfect squares can have an odd number of divisors.

The dragon goes over all the triangle numbers, i.e., numbers that are the sum of consecutive integers from 1 to m.

Our challenge was to find an integer n that satisfies

n = 576+1000*l = k*k = m*(m+1)/2

for some integers l, k, and m. These numbers satisfy ( http://en.wikipedia.org/wiki/Square_triangular_number) a recursion formula

n_{k+1} = 34*n_k - n_{k-1} + 2 with the initial value of n_1=1 and n_0=0.

Since we are interested in n modulo (7*24*60*60), we can compute the full cycle, using the recursion above, and get three possible solutions:

Thu 12:09:36; Mon 21:09:36; and Mon 00:09:36.

If you have any problems you think we might enjoy, please send them in. All replies should be sent to: ponder@il.ibm.com