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
You want to prove (q ∨ r) → p by contrapositive. Which implication do you prove instead? Click it.
(q ∨ r) → p
1¬p → ¬(q ∨ r)
2(q ∨ r) ∧ ¬p
3p → (q ∨ r)
4¬(q ∨ r) → ¬p
2.Propositional logic & proofsHard
The premises a → c; ¬c are given. Which conclusion follows from them? Click it.
a → c¬c
1¬a
2c
3a
4¬a ∧ c
3.Sets, relations & functionsMedium
In a group of 52 students, 24 speak French (F), 16 speak German (G) and 23 speak Spanish (S). 9 speak French and German, 11 French and Spanish, 4 German and Spanish, and 1 speak all three. How many are in exactly two of the three sets? Call the number n.
|U| = 52
n =
4.Sets, relations & functionsHard
An equivalence relation R on {1, …, 10} has the classes {2, 9}, {4, 5, 6, 8}, {1, 10}, {3, 7}. How many ordered pairs does R contain? Call the number |R|.
{2, 9}{4, 5, 6, 8}{1, 10}{3, 7}
|R| =
5.Graph theoryMedium
A tree has 4 vertices of degree 3, 4 vertices of degree 5, 3 vertices of degree 2, and all its other vertices are leaves (degree 1). How many leaves L does it have?
4 × deg 34 × deg 53 × deg 2
L =
6.Graph theoryHard
Find the total weight W of a minimum spanning tree of the weighted graph.
|V| = 7|E| = 9
W =
7.Number theory & modular arithmeticMedium
Solve 39x ≡ 38 (mod 89). Give the solution x with 0 < x < 89.
39x ≡ 38 (mod 89)
x =
8.Number theory & modular arithmeticHard
An RSA key uses the primes p = 11 and q = 41 and the public exponent e = 19. Find the private exponent d (0 < d < φ(n)).
p = 11q = 41e = 19
d =
3Solutions
1.Propositional logic & proofs
Answer: ¬p → ¬(q ∨ 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).
Parts of the implication
a = q ∨ r, b = p
Contrapositive
¬b → ¬a = ¬p → ¬(q ∨ r)
Answer
¬p → ¬(q ∨ r)
The contrapositive is equivalent to the original statement, so proving it proves the statement.
2.Propositional logic & proofs
Answer: ¬a
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
a = 0, c = 0: ¬a = 1, c = 0, a = 0, ¬a ∧ c = 0
Conclusion
¬a (modus tollens)
Answer
¬a
3.Sets, relations & functions
Answer: n = 21
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
F ∩ G ∩ S: 1
F ∩ G only: 9 − 1 = 8
F ∩ S only: 11 − 1 = 10
G ∩ S only: 4 − 1 = 3
Single sets
F only: 24 − 8 − 10 − 1 = 5
G only: 16 − 8 − 3 − 1 = 4
S only: 23 − 10 − 3 − 1 = 9
Answer
8 + 10 + 3 = 21
Answer
n = 21
4.Sets, relations & functions
Answer: |R| = 28
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.
Pairs per class
|R| = 2² + 4² + 2² + 2² = 28
Answer
|R| = 28
5.Graph theory
Answer: L = 18
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 + 38 = 2(L + 11) − 2
L = 18
Answer
L = 18
Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.
6.Graph theory
Answer: W = 53
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
AB (5): take
GD (7): take
EF (8): take
GE (9): take
CE (10): take
DF (12): skip, closes a cycle
BD (14): take
Total weight
W = 5 + 7 + 8 + 9 + 10 + 14 = 53
Answer
W = 53
7.Number theory & modular arithmetic
Answer: x = 74
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 39
r = 89, t = 0
r = 39, q = 2, t = 1
r = 11, q = 3, t = −2
r = 6, q = 1, t = 7
r = 5, q = 1, t = −9
r = 1, q = 5, t = 16
39−1 ≡ 16 (mod 89)
Solve
x ≡ 16·38 = 608 ≡ 74 (mod 89)
check: 39·74 = 2886 ≡ 38 (mod 89)
Answer
x = 74
8.Number theory & modular arithmetic
Answer: d = 379
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.