Draw a Venn diagram and decide region by region what belongs to the set expression.
For counting problems, fill in the regions from the centre outwards: all sets, then pairs, then single sets, then the outside.
Use inclusion–exclusion as a check: |A ∪ B| = |A| + |B| − |A ∩ B|.
For a relation, test each property on every pair and give one counterexample when it fails.
For a function, count the arrows arriving at each element of B: at most one means injective, at least one surjective.
Graph theory
Handshake lemma
Σ deg(v) = 2|E|
Complete graphs
|E(Kn)| = n(n − 1)/2, |E(Km,n)| = m·n
Regular graph
|E| = n·k/2
Tree and forest
|E| = n − 1, |E| = n − c
Leaves of a tree
L = 2 + Σ (deg(v) − 2) over deg(v) ≥ 3
Euler
circuit: all degrees even; path: exactly 2 odd degrees
Hamilton circuits in Kn
(n − 1)!/2
Dijkstra update
d(v) = min(d(v), d(u) + w(u, v))
Walks of length L
(Mᴸ)ij
Method
Write down the degree of every vertex; the degrees add up to twice the number of edges.
For Euler questions, check that the graph is connected and count the vertices with odd degree.
For trees, use n − 1 edges and the handshake lemma.
For Dijkstra, keep a table of tentative distances and settle the smallest unsettled vertex each round.
For a minimum spanning tree, sort the edges by weight and skip every edge that closes a cycle (Kruskal), or grow one tree from a start vertex (Prim).
Number theory & modular arithmetic
Division with remainder
a = q·n + r, 0 ≤ r < n
Euclid
gcd(a, b) = gcd(b, a mod b)
Least common multiple
lcm(a, b) = a·b/gcd(a, b)
Inverse
a·a⁻¹ ≡ 1 (mod n), gcd(a, n) = 1
Linear congruence
a·x ≡ b → x ≡ a⁻¹·b (mod n)
Euler’s totient
φ(n) = n·∏(1 − 1/p), φ(pᵏ) = pᵏ − pᵏ⁻¹
Fermat
aᵖ⁻¹ ≡ 1 (mod p)
RSA
n = pq, φ = (p − 1)(q − 1), d = e⁻¹ mod φ, c = mᵉ mod n, m = cᵈ mod n
Method
Reduce every number modulo n first; remainders lie between 0 and n − 1.
For a gcd, divide repeatedly with Euclid’s algorithm; the last nonzero remainder is the gcd.
For an inverse, run the extended algorithm and keep the coefficient of a; add n if it is negative.
Solve a·x ≡ b by multiplying with a⁻¹; combine two congruences by writing x = r₁ + m₁k.
For powers, write the exponent in binary and square repeatedly; in RSA, d = e⁻¹ mod φ(n) with φ(n) = (p − 1)(q − 1).
2Practice problems
1.Propositional logic & proofsMedium
Is r ↔ ¬r a tautology, a contradiction or a contingency? Click your answer.
r ↔ ¬r
1tautology
2contradiction
3contingency
2.Propositional logic & proofsHard
The premises p ∨ q; ¬p are given. Which conclusion follows from them? Click it.
p ∨ q¬p
1p ∧ q
2¬q
3q
4p
3.Sets, relations & functionsMedium
Of 58 first-year students, 27 passed Mathematics (M), 25 passed Physics (P) and 23 passed Chemistry (C). 13 passed M and P, 8 passed M and C, 9 passed P and C, and 5 passed all three. How many are in none of the three sets?
|U| = 58
|(M ∪ P ∪ C)′| =
4.Sets, relations & functionsHard
How many different relations are there on a set with 4 elements? Call the number N.
|X| = 4
N =
5.Graph theoryMedium
Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.
|V| = 7|E| = 11
1Euler circuit
2Euler path, no circuit
3neither
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
Solve 17x ≡ 12 (mod 31). Give the solution x with 0 < x < 31.
17x ≡ 12 (mod 31)
x =
8.Number theory & modular arithmeticHard
Encrypt the message m = 86 with the RSA public key (n, e) = (341, 23). Find the ciphertext c.
n = 341e = 23m = 86
c =
3Solutions
1.Propositional logic & proofs
Answer: contradiction
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).
Truth table row by row
r = 1: ¬r = 0, r ↔ ¬r = 0
r = 0: ¬r = 1, r ↔ ¬r = 0
Classify
r ↔ ¬r: 0, 0
0 in every row: a contradiction.
Answer
contradiction
2.Propositional logic & proofs
Answer: q
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.
Rows with all premises true
p = 0, q = 1: p ∧ q = 0, ¬q = 0, q = 1, p = 0
Conclusion
q (disjunctive syllogism)
Answer
q
3.Sets, relations & functions
Answer: |(M ∪ P ∪ C)′| = 8
Sets and Venn diagrams: A ∪ B holds the elements in A or B (or both), A ∩ B the elements in both, A − B the elements of A that are not in B, and the complement A′ everything in U that is not in A. In a Venn diagram every set expression is a union of regions, so count it region by region. Inclusion–exclusion: |A ∪ B| = |A| + |B| − |A ∩ B|, and |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C|.
Centre and pairs
M ∩ P ∩ C: 5
M ∩ P only: 13 − 5 = 8
M ∩ C only: 8 − 5 = 3
P ∩ C only: 9 − 5 = 4
Single sets
M only: 27 − 8 − 3 − 5 = 11
P only: 25 − 8 − 4 − 5 = 8
C only: 23 − 3 − 4 − 5 = 11
Answer
|M ∪ P ∪ C| = 50
|(M ∪ P ∪ C)′| = 58 − 50 = 8
Answer
|(M ∪ P ∪ C)′| = 8
4.Sets, relations & functions
Answer: N = 65536
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.
Count the free choices
N = 216 = 65536
Answer
N = 65536
5.Graph theory
Answer: Euler path, no circuit
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.
The Euler path starts at one odd vertex and ends at the other.
6.Graph theory
Answer: d(F) = 10
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: 8, C: 4
C (4): B: 7, E: 9, G: 5
G (5): E: 8
B (7)
E (8): D: 15, F: 10
F (10): D: 11
D (11)
Shortest path
A → C → G → E → F
d(F) = 10
Answer
d(F) = 10
7.Number theory & modular arithmetic
Answer: x = 8
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).
Inverse of 17
r = 31, t = 0
r = 17, q = 1, t = 1
r = 14, q = 1, t = −1
r = 3, q = 4, t = 2
r = 2, q = 1, t = −9
r = 1, q = 2, t = 11
17−1 ≡ 11 (mod 31)
Solve
x ≡ 11·12 = 132 ≡ 8 (mod 31)
check: 17·8 = 136 ≡ 12 (mod 31)
Answer
x = 8
8.Number theory & modular arithmetic
Answer: c = 300
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.