Algebra A Solutions 1. We have a32 + a33 + a34 = 4, a35 + a36 + a37 = 7, a38 + a39 + a310 = 10, and a31 = 8000. Thus, the answer is 8021 . Problem written by Eric Neyman. 2. Note that f (f (f (f (f (x))))) = f (x) is a linear equation in x and thus has one solution. If f (x) = x then clearly this equation is satisfied. Thus the solution is the solution to the equation x = 15x − 2016, which is 144 . Problem written by Eric Neyman. 3. We have 4096f (f (x, x), x) = x13 log2 x 4096 xlog2 x = x13 2 4096x(log2 x) = x13 2 4096x(log2 x) −13 =1 (log2 x)((log2 x)2 −13)+12 = 20 2 (log2 x)((log2 x)2 − 13) + 12 = 0 (log2 x)3 − 13 log2 x + 12 = 0 (log2 x − 1)(log2 x − 3)(log2 x + 4) = 0, so x can be any of 2, 8, and answer is 161 + 16 = 177 . 1 16 . Thus, the sum of all possible values of x is 161 16 , and so our Problem written by Eric Neyman. 4. Using the well known result that x1 −x2 | P (x1 )−P (x2 ) we get that N ≡ 2 (mod 2015), N ≡ 3 (mod 2014) and N ≡ 3 (mod 2013). Solving these equations gives N ≡ 3 + 1007 × 2013 × 2014 (mod 2013 × 2014 × 2015). Thus N = 3 + 1007 × 2013 × 2014 and N ≡ 3 + 1007 × (−3) × (−2) ≡ 2013 (mod 2016). The polynomial P (x) = (x − 1)((x − 2)(1006) + 1) + 2 satisfies all given conditions, so 2013 is the final answer. Problem written by Mel Shu. 5. By a computation, a5 = 212 −1. If ai = 2k −1, then ai+1 = (2k−1 −1)2k+1 , so ai+k+2 = 2k−1 −1. Eventually we get a105 = 0. Thus the answer is 105 . Problem written by Zhuo Qun Song. 6. We can write [a, b] = (a − 1)(b − 1) − 1. Since [[a, b], [c, d]] ≤ [[[a, b], c], d] if a > b > c > d, we can maximize V by finding [[· · · [[101, 100], 99], · · · ], 2] = 101 Y (k − 1) − 2 j 99 Y X (k − 1) − (2 − 1) − 1. j=2 k=2 k=2 This is perhaps most easily seen by evaluating [[[a, b], c], d] = (a − 1)(b − 1)(c − 1)(d − 1) − 2(c − 1)(d − 1) − 2(d − 1) − 1 1 and generalizing from there; the general form is [[· · · [[an , an−1 ], a1 ], · · · ], 2] = n Y (ak − 1) − 2 n−2 X j Y (ak − 1) − (a1 − 1) − 1. j=1 k=1 k=1 Clearly the first term, 100!, dominates, so N = 100. So, the desired value becomes 2 · 1 1 1 + 100·99·98 + 100·99·98·97 + · · · ) as the remaining terms in the sum are far too small to 106 ( 100·99 1 1 + 99·98 ). change our value. In fact, 100·99·98·97 >> 2·106 , so we actually wish to find 2·104 ( 99 This is just a smidge over 204 . Problem written by Zachary Stier. Q 7. We first determine all solutions to the equation P (x2 ) = P (x)P Q (x2− 1). Let P Q(x) = (xQ− αi ) where αi are the roots of P (including multiplicity). Then (x − αi ) = (x − αi ) (x − √ (αi + 1)). Thus, considering the sets of roots of both sides of this equation, we get {±2 αi } = √ {αi } ∪ {αi + 1}. Now we note that if max αi = M > 1 then max |αi | = M > M , √ contradiction. Similarly, if min αi = N < 1 then min |αi | = N 2 < N , contradiction. (Unless N = 0. But this is impossible, since the multiplicity of zero as a root on both sides of the equation cannot be equal.) Therefore all roots αi must satisfy |αi | = |αi + 1| = 1, so αi = ± √ −1+ 3i 2 . Now counting the distinct roots, the two different roots must have the same multiplicity, so the solutions are P (x) = (x2 + x + 1)n for some positive integer n, or P (x) = 0 or P (x) = 1 (constant solutions). It follows that P0 (1) = 3n , so P0 (x) = (x2 +x+1)2 . Therefore P0 (10) = 1112 = 12321 . Problem written by Mel Shu. 8. For |x| < 1, we have: a b f (x) − f (x−1 ) = a b x2 3 x−2 3 + 1 − x2a+1 3b+1 x−2a+1 3b+1 − 1 a,b≥0 X X X x2a 3b + x5·2a 3b = 1 − x2a+1 3b+1 a≥0 b≥0 X X a 2u+1 = x2 a≥0 u≥0 = X xv v>0 = x 1−x Similarly, for |x| > 1, we have: f (x) = f (x−1 ) = Since the range of Setting y 1−y 1 1−x 1 . 1−x over the domain |x| > 1 is (−∞, 0) ∪ (0, 12 ), it follows that |y| < 1. = 2016 yields y = 2016 2017 , so p + q = 4033 . Problem written by Bill Huang. If you believe that any of these answers is incorrect, or that a problem had multiple reasonable interpretations or was incorrectly stated, you may appeal at tinyurl.com/pumacappeals. All appeals must be in by 1 PM to be considered. 2

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