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
Which formula is logically equivalent to ¬(q → p)? Click it.
¬(q → p)
1¬q → ¬p
2¬q ∨ ¬p
3q → ¬p
4q ∧ ¬p
2.Propositional logic & proofsHard
Which statement is equivalent to ¬∃x (P(x) ∧ Q(x)), with the negation moved all the way inside? Click it.
¬∃x (P(x) ∧ Q(x))
1∀x (P(x) ∧ ¬Q(x))
2∀x (¬P(x) ∧ ¬Q(x))
3∀x (P(x) → ¬Q(x))
4∃x (P(x) → ¬Q(x))
3.Sets, relations & functionsMedium
Exactly one region of the Venn diagram is the set (B ∩ C) ∩ (B′ ∪ A). Click its number.
(B ∩ C) ∩ (B′ ∪ A)
1region 1
2region 2
3region 3
4region 4
5region 5
6region 6
7region 7
8region 8
4.Sets, relations & functionsHard
The bijection f on {1, 2, 3, 4, 5} is given by its table. Find f−1(2).
f: 1 ↦ 4, 2 ↦ 3, 3 ↦ 2, 4 ↦ 1, 5 ↦ 5
f−1(2) =
5.Graph theoryMedium
A tree has 4 vertices of degree 3, 2 vertices of degree 4, 4 vertices of degree 5, and all its other vertices are leaves (degree 1). How many leaves L does it have?
4 × deg 32 × deg 44 × deg 5
L =
6.Graph theoryHard
Find the total weight W of a minimum spanning tree of the weighted graph.
|V| = 7|E| = 10
W =
7.Number theory & modular arithmeticMedium
Find the inverse of 31 modulo 87: the number x with 0 < x < 87 and 31·x ≡ 1 (mod 87).
a = 31n = 87
31−1 mod 87 =
8.Number theory & modular arithmeticHard
Encrypt the message m = 185 with the RSA public key (n, e) = (187, 13). Find the ciphertext c.
n = 187e = 13m = 185
c =
3Solutions
1.Propositional logic & proofs
Answer: q ∧ ¬p
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
¬(a → b) ≡ ¬(¬a ∨ b) ≡ a ∧ ¬b: the only row where an implication fails.
¬(q → p) ≡ q ∧ ¬p
Check with the truth table
p = 1, q = 1: 0 | 1 0 0 0
p = 1, q = 0: 0 | 0 1 1 0
p = 0, q = 1: 1 | 1 1 1 1
p = 0, q = 0: 0 | 1 1 1 0
Answer
q ∧ ¬p
The columns after the bar are the options in the order shown.
2.Propositional logic & proofs
Answer: ∀x (P(x) → ¬Q(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 (P(x) ∧ Q(x)) ≡ ∀x (P(x) → ¬Q(x))
Answer
∀x (P(x) → ¬Q(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: region 4
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|.
Inside or outside each set
A ∩ B ∩ C
The region
region 4
Answer
region 4
4.Sets, relations & functions
Answer: f−1(2) = 3
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.
Read the table backwards
f(3) = 2 → f−1(2) = 3
Answer
f−1(2) = 3
5.Graph theory
Answer: L = 22
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 + 40 = 2(L + 10) − 2
L = 22
Answer
L = 22
Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.
6.Graph theory
Answer: W = 33
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.
Kruskal’s algorithm
BC (1): take
AB (2): take
CE (3): take
DF (5): take
DE (6): take
BD (15): skip, closes a cycle
GC (16): take
Total weight
W = 1 + 2 + 3 + 5 + 6 + 16 = 33
Answer
W = 33
7.Number theory & modular arithmetic
Answer: 31−1 mod 87 = 73
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·31 (mod 87).
r = 87, t = 0
r = 31, q = 2, t = 1
r = 25, q = 1, t = −2
r = 6, q = 4, t = 3
r = 1, q = 6, t = −14
Inverse
1 ≡ −14·31 (mod 87)
x = 73
check: 31·73 = 2263 = 26·87 + 1
Answer
31−1 mod 87 = 73
8.Number theory & modular arithmetic
Answer: c = 36
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.