How many surjective (onto) functions f: A → B are there if |A| = 4 and |B| = 3? Call the number N.
|A| = 4|B| = 3
N =
5.Graph theoryMedium
A tree has 2 vertices of degree 3, 3 vertices of degree 2, and all its other vertices are leaves (degree 1). How many leaves L does it have?
2 × deg 33 × deg 2
L =
6.Graph theoryHard
Use Dijkstra’s algorithm to find the length d(F) of the shortest path from A to F in the weighted graph.
start: Aend: F
d(F) =
7.Number theory & modular arithmeticMedium
Find the inverse of 40 modulo 51: the number x with 0 < x < 51 and 40·x ≡ 1 (mod 51).
a = 40n = 51
40−1 mod 51 =
8.Number theory & modular arithmeticHard
Use repeated squaring to find 17127 mod 67.
a = 17k = 127n = 67
17127 mod 67 =
3Solutions
1.Propositional logic & proofs
Answer: ¬p ∨ ¬r
Logical equivalence: two formulas are equivalent (≡) when they have the same value in every row of the truth table. A tautology is 1 in every row, a contradiction is 0 in every row, and a contingency is neither. Useful laws: De Morgan, ¬(a ∧ b) ≡ ¬a ∨ ¬b and ¬(a ∨ b) ≡ ¬a ∧ ¬b; a → b ≡ ¬a ∨ b ≡ ¬b → ¬a (contrapositive); ¬(a → b) ≡ a ∧ ¬b; and the distributive laws, a ∨ (b ∧ c) ≡ (a ∨ b) ∧ (a ∨ c).
The law
De Morgan: ¬(a ∧ b) ≡ ¬a ∨ ¬b. The ∧ turns into ∨ and both parts are negated.
¬(p ∧ r) ≡ ¬p ∨ ¬r
Check with the truth table
p = 1, r = 1: 0 | 0 0 0 1
p = 1, r = 0: 1 | 0 1 0 0
p = 0, r = 1: 1 | 1 1 0 1
p = 0, r = 0: 1 | 0 1 1 1
Answer
¬p ∨ ¬r
The columns after the bar are the options in the order shown.
2.Propositional logic & proofs
Answer: ∀x ¬A(x)
Quantifiers and arguments: ¬∀x P(x) ≡ ∃x ¬P(x) and ¬∃x P(x) ≡ ∀x ¬P(x). Move a negation inwards, swap every quantifier it passes, and negate the inner formula with the propositional laws, such as ¬(a → b) ≡ a ∧ ¬b. An argument is valid when the conclusion is 1 in every row in which all premises are 1. A single row with true premises and a false conclusion is a counterexample and shows the argument is invalid.
Move the negation inwards
¬∃x A(x) ≡ ∀x ¬A(x)
Answer
∀x ¬A(x)
The other options either keep a quantifier unchanged, change the order of the quantifiers, or negate the inner formula wrongly.
3.Sets, relations & functions
Answer: transitive
Relations: a relation R on a set X is a set of ordered pairs (x, y). R is reflexive when (x, x) ∈ R for every x in X; symmetric when (x, y) ∈ R always comes with (y, x) ∈ R; antisymmetric when (x, y) and (y, x) are both in R only if x = y; and transitive when (x, y) and (y, z) in R force (x, z) ∈ R. An equivalence relation is reflexive, symmetric and transitive, and splits X into disjoint classes.
reflexive
fails: (3, 3) ∉ R
symmetric
fails: (1, 3) ∈ R, (3, 1) ∉ R
antisymmetric
fails: (1, 2), (2, 1) ∈ R
transitive
holds
Answer
transitive
4.Sets, relations & functions
Answer: N = 36
Functions and counting: a function f: A → B gives every element of A exactly one image in B. f is injective (one-to-one) when different elements have different images, surjective (onto) when every element of B is an image, and bijective when both hold. With |A| = m and |B| = n there are nᵐ functions, n!/(n − m)! injective ones (m ≤ n), and by inclusion–exclusion Σ (−1)ᵏ·C(n, k)·(n − k)ᵐ surjective ones. |𝒫(A)| = 2ᵐ and |A × B| = m·n.
Count
N = C(3, 0)·34 − C(3, 1)·24 + C(3, 2)·14 = 36
Answer
N = 36
5.Graph theory
Answer: L = 4
Euler, Hamilton and trees: a connected graph has an Euler circuit (every edge once, back to the start) when every vertex has even degree, and an Euler path that is not a circuit when exactly two vertices have odd degree; the path then runs from one odd vertex to the other. A Hamilton circuit visits every vertex once; Kn has (n − 1)!/2 of them. A tree on n vertices has n − 1 edges, and a connected graph with m edges keeps a spanning tree after removing m − n + 1 edges.
Handshake lemma for a tree
L + 12 = 2(L + 5) − 2
L = 4
Answer
L = 4
Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.
6.Graph theory
Answer: d(F) = 15
Shortest paths and spanning trees: Dijkstra’s algorithm settles one vertex per round: the unsettled vertex with the smallest tentative distance. Its distance is then final, and every neighbour v gets d(v) = min(d(v), d(u) + w(u, v)). Kruskal’s algorithm builds a minimum spanning tree by taking the edges from light to heavy and skipping any edge that would close a cycle; Prim’s algorithm grows one tree from a start vertex by always adding the lightest edge that leaves it. A spanning tree of n vertices has n − 1 edges.
Dijkstra round by round
Each line: the vertex settled, its final distance, and the neighbours whose distance improves.
A (0): B: 10, C: 15
B (10): D: 12, G: 23
D (12): E: 15, F: 15
C (15)
E (15): G: 22
F (15)
G (22)
Shortest path
A → B → D → F
d(F) = 15
Answer
d(F) = 15
7.Number theory & modular arithmetic
Answer: 40−1 mod 51 = 37
Inverses and congruences: a has an inverse modulo n exactly when gcd(a, n) = 1. The extended Euclidean algorithm writes 1 = s·n + t·a, and then a⁻¹ ≡ t (mod n). A congruence a·x ≡ b (mod n) with gcd(a, n) = 1 has the solution x ≡ a⁻¹·b. Two congruences x ≡ r₁ (mod m₁) and x ≡ r₂ (mod m₂) with coprime moduli have one solution modulo m₁·m₂ (Chinese remainder theorem). Euler’s φ(n) counts the numbers from 1 to n that are coprime to n: φ(n) = n·∏(1 − 1/p).
Extended Euclidean algorithm
Each remainder r is written as r ≡ t·40 (mod 51).
r = 51, t = 0
r = 40, q = 1, t = 1
r = 11, q = 3, t = −1
r = 7, q = 1, t = 4
r = 4, q = 1, t = −5
r = 3, q = 1, t = 9
r = 1, q = 3, t = −14
Inverse
1 ≡ −14·40 (mod 51)
x = 37
check: 40·37 = 1480 = 29·51 + 1
Answer
40−1 mod 51 = 37
8.Number theory & modular arithmetic
Answer: 17127 mod 67 = 19
Modular powers and RSA: compute aᵏ mod n by repeated squaring: write k in binary, square, reduce after every step and multiply the powers whose bit is 1. Fermat: aᵖ⁻¹ ≡ 1 (mod p) for a prime p not dividing a, so the exponent can be reduced modulo p − 1. RSA: choose primes p and q, n = p·q and φ(n) = (p − 1)(q − 1); the public exponent e is coprime to φ(n), and the private exponent is d = e⁻¹ mod φ(n). Encrypt with c = mᵉ mod n and decrypt with m = cᵈ mod n.