About Last Month
1. Find the last three digits of \(2011^{2012}\).
The first thing you might notice about this problem which can make it significantly easier is that \(2011\equiv 11 \pmod {1000}\) so we can consider the simpler problem of finding the last three digits of \(11^{2012}\). From here, I used the binomial theorem to write
$$11^{2012} = (10+1)^{2012} = \sum_{i=0}^{2012} \binom{2012}{i} 10^{i}1^{2012-i} $$
Notice that in the sum, anything with a \(10^3\) or higher will end in all zeros, so the last three digits are the last three digits of
$$\binom{2012}{2}10^2 + \binom{2012}{1}10 + \binom{2012}{0} = 202,306,600 + 20,120 +1$$
from which we can see the last 3 digits are 721.
2. Evaluate \(\sqrt{1 + \sqrt{ 1 + \sqrt{ 1 + \cdots } } } \)
Let \(x=\sqrt{1 + \sqrt{ 1 + \sqrt{ 1 + \cdots } } } \). Then we see that \(x=\sqrt{1+x}\) from which we get
$$x=\frac{1\pm\sqrt5}{2}$$
since \((1-\sqrt 5)/2\) is negative, the answer must be \((1+\sqrt 5)/2\) which is the golden ratio. Interestingly, the golden ratio is also equal to the continued fraction
$$1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \ddots}}}$$
and the solution can be obtained in a very similar manner (in this case, \(x=1+\frac 1x\)).
The Problem
I'm low on inspiration this month, so I'm stealing another one for the Citadel's problem of the week. I think there are probably a lot of ways to approach this one, so I'm curious to see what people can come up with!
1. Lagrange proved that any positive integer can be written as the sum of four squares. For instance, \(19=3^2+3^2+1^2+0^2\) and \(97=6^2+6^2+5^2+0^2\). Express 2012 as the sum of four squares in two different ways--first, as the sum of squares where one number appears twice and second where all 4 numbers are different.
2. http://spikedmath.com/puzzle-006.html (I haven't worked this one out yet, so I don't know the answer!)
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