1 . 7 ( U/L ) th Answer: 1231288 For x to be such a number is equivalent to x being an k root of unity for some th k up to 2012. For each k , there are ϕ ( k ) primitive k roots of unity, so the total number of roots is ∑ 2012 ϕ ( k ). k =1 We will give a good approximation of this number using well known facts about the M¨ obius function, { ∑ 0 if n is not squarefree defined by μ ( n ) = . It turns out that if f ( n ) = g ( d ), d | n r ( − 1) if n has r distinct prime factors ∑ ∑ ∑ n n then g ( n ) = μ ( d ) f ( ). Using this fact, since n = ϕ ( d ), we have that ϕ ( n ) = μ ( d ) . d | n d | n d | n d d ∑ ∑ ∑ ∑ 2012 2012 k k Now we have reduced the problem to estimating μ ( d ) . Let a = , so we obtain aμ ( d ). k =1 d | k k =1 d | k d d Guts We can interchange the order of summation by writing 2012 ⌊ ⌋ ( ) 2012 2012 2 d ∑ ∑ ∑ 1 2012 aμ ( d ) ≈ μ ( d ) ⌊ ⌋ 2 d a =1 d =1 d =1 2012 2 ∑ 2012 ≈ μ ( d ) 2 2 d d =1 2012 2 ∑ 2012 μ ( d )

2 2 d d =1 ∞ 2 ∑ 2012 μ ( d ) ≈ 2 2 d d =1 { ∑ 1 if n = 1 The M¨ obius function also satisfies the property that μ ( d ) = , which can be seen d | n 0 otherwise { 1 if n = 1 as a special case of the theorem above (letting f ( n ) = 1 , g ( n ) = ). We can then see that 0 otherwise ( ) (∑ ) ∑ ∑ ∑ ∞ μ ( d ) ∞ ∞ μ ( d ) 2012 1 1 6 3 2 = = 1, so = . Therefore, we have ϕ ( k ) ≈ · 2012 = 2 2 2 2 2 2 d =1 c =1 d =1 k =1 d c 1 d π π 1230488 . 266... 2012 is large enough that all of our approximations are pretty accurate and we should be comfortable perturbing this estimate by a small factor to give bounding values.