Math 522 Exam 1 Solutions
1. Youre playing Fibonacci nim. You start with 66, and you go rst.
What are all your possible (initial) winning moves?
We write 66 = 55 + 8 + 3, as the sum of nonconsecutive
Fibonacci numbers. We may take 3, leaving our opponent
KEANE 'CRYSTAL BALL'
Who is the man I see
Where Im supposed to be?
I lost my heart, I buried it too deep
Under the iron sea
Oh, crystal ball, crystal ball
Save us all, tell me life is beautiful
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Math 522 Exam 13 Solutions
1. Let a, b, c Z. Suppose that b, c are odd and positive. Prove that
a
( bc ) = ( a )( a ).
b
c
We factor b = p1 p2 pk into (not necessarily distinct) odd
primes. Similarly, we factor c = q1 q2 qm . Now bc =
p1 p2 pk q1 q2 qm .
Math 522 Exam 12 Solutions
1. Let n N. Suppose we have players P1 , P2 , . . . , Pn , with additive valuation functions
f1 , f2 , . . . , fn . Let B1 , B2 , . . . , Bn be bundles of goods and/or payments. Suppose that
we assign bundle Bi to player Pi (for
Math 522 Exam 11 Solutions
1. For n = 13, a = 2 is a primitive root, whose index table is below
for convenience. Use indices to nd all solutions to the congruence
987
654
5x4 x11 (mod 13).
k 1
2k 2
2
4
3
8
987
4
3
5
6
6
12
7 8 9 10 11
11 9 5 10 7
12
1
654
Math 522 Exam 10 Solutions
1. Prove that r|n (r) = 2s , where s is the number of dierent primes
d(r)
dividing n, i.e. n = pa1 pa2 pas .
1 2
s
Set G(n) = r|n (r) . Note that is multiplicative, since
d(r)
d
both the numerator and denominator are (and the de
Math 522 Exam 9 Solutions
Theorem 1. Let m, n N. If gcd(m, n) = 1 then (mn) = (m)(n).
Theorem 2. Let p, k N. If p is prime, then (pk ) = pk pk1 .
1. Use the two theorems above to prove the following:
1
1 p .
Claim. For all n N, (n) = n
p|n
Let n = pk1 pk2
Math 522 Exam 4 Solutions
1. Prove the converse of Wilsons theorem. That is, suppose that p > 1 is not prime.
Prove that p (p 1)! + 1.
Since p is not prime, it is not irreducible, so we may write p = mn where
1 < m < p. Because m p 1, we have m|(p 1)!. Su
Math 522 Exam 8 Solutions
1. Find all
x
x
x
solutions to the following system of congruences:
3 (mod 6)
7 (mod 10)
12 (mod 15)
We rst reduce the congruences as: x 3 (mod 2), x 3 (mod 3), x
7 (mod 2), x 7 (mod 5), x 12 (mod 3), x 12 (mod 5). Simplifyin
Math 522 Exam 6 Solutions
1. Suppose that a, b Z and k N. Suppose that a b (mod k), |a| < k , and |b|
3
Prove that a = b.
2k
.
3
Because a b, there is some n Z with a b = nk. For convenience, we
rewrite our inequalities as k < a < k and 2k b 2k .
3
3
3
3
Math 522 Exam 2 Solutions
1. Consider the ring R[x], polynomials with real coecients. Two elements from this ring are f (x) = x4 2x2 8, g(x) = x3 + x2 + 2x + 2.
Use the Euclidean algorithm to nd gcd(f (x), g(x).
We use long division to nd q1 (x), r1 (x) s
Math 522 Exam 7 Solutions
1. Find a, the inverse of a modulo c, when a = 10 and c = 123.
We will use the Euclidean algorithm to nd a solution to
10x + 123y = 1; then x will be the answer we seek. First,
we have 123 = 12 10 + 3. Second, we have 10 = 3 3 +
Math 522 Exam 5 Solutions
1. Let c0 = 0, c1 = 1, c2 = 1, c3 = 2, c4 = 5, . . . denote the Catalan numbers. Let
n
sn =
i=0 ci denote the sum of the rst n Catalan numbers. Find a generating
function S(x) for the sequence cfw_sn .
1 14x
1
n
Let A(x) = 1x =
i
Math 522 Exam 3 Solutions
1. For all k N, nd all real solutions x to the equation
x
k
=
x
k+1
.
We rst use the denition of the binomial coecient to get
1 k
1
1
x = k! x(x 1) (x k + 1) = (k+1)! x(x 1) (x
k!
k + 1)(x k). Multiplying by (k + 1)! on both sid
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Juan Lu