### Algebra A - PUMaC 2016

Algebra A Solutions
1. We have a32 + a33 + a34 = 4, a35 + a36 + a37 = 7, a38 + a39 + a310 = 10, and a31 = 8000. Thus, the
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.
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