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 c → ¬b by contrapositive. Which implication do you prove instead? Click it.
c → ¬b
1c ∧ ¬¬b
2¬c → ¬¬b
3¬¬b → ¬c
4¬b → c
2.Propositional logic & proofsHard
The argument with premises a → c; c and conclusion a is invalid. Click the row of the truth table that is a counterexample.
a → cc∴ a
1a = 1, c = 1
2a = 1, c = 0
3a = 0, c = 1
4a = 0, c = 0
3.Sets, relations & functionsMedium
A survey of 55 people finds that 24 own a laptop (L), 23 a tablet (T) and 35 a smartwatch (W). 8 own L and T, 13 own L and W, 15 own T and W, and 1 own all three. How many are in exactly one of the three sets? Call the number n.
|U| = 55
n =
4.Sets, relations & functionsHard
On X = {1, 2, …, 15} the relation x ~ y holds when x ≡ y (mod 4). How many elements are in the equivalence class [8]?
X = {1, …, 15}x ~ y ⟺ 4 | (x − y)
|[8]| =
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| = 6|E| = 8
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
Find the inverse of 3 modulo 53: the number x with 0 < x < 53 and 3·x ≡ 1 (mod 53).
a = 3n = 53
3−1 mod 53 =
8.Number theory & modular arithmeticHard
Use repeated squaring to find 19107 mod 69.
a = 19k = 107n = 69
19107 mod 69 =
3Solutions
1.Propositional logic & proofs
Answer: ¬¬b → ¬c
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 = c, b = ¬b
Contrapositive
¬b → ¬a = ¬¬b → ¬c
Answer
¬¬b → ¬c
The contrapositive is equivalent to the original statement, so proving it proves the statement.
2.Propositional logic & proofs
Answer: a = 0, c = 1
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.
Premises and conclusion row by row
a = 1, c = 1: 1 1 | 1
a = 1, c = 0: 0 0 | 1
a = 0, c = 1: 1 1 | 0
a = 0, c = 0: 1 0 | 0
Counterexample
a = 0, c = 1
Answer
a = 0, c = 1
An argument with this form looks like a valid rule, but one row is enough to show that it is not.
3.Sets, relations & functions
Answer: n = 13
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
L ∩ T ∩ W: 1
L ∩ T only: 8 − 1 = 7
L ∩ W only: 13 − 1 = 12
T ∩ W only: 15 − 1 = 14
Single sets
L only: 24 − 7 − 12 − 1 = 4
T only: 23 − 7 − 14 − 1 = 1
W only: 35 − 12 − 14 − 1 = 8
Answer
4 + 1 + 8 = 13
Answer
n = 13
4.Sets, relations & functions
Answer: |[8]| = 3
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.
The class
[8] = {4, 8, 12}
|[8]| = 3
Answer
|[8]| = 3
5.Graph theory
Answer: neither
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.
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: 1
C (1): B: 9, E: 14
B (9): D: 23, E: 13
E (13): D: 22, F: 21
F (21)
D (22)
Shortest path
A → C → B → E → F
d(F) = 21
Answer
d(F) = 21
7.Number theory & modular arithmetic
Answer: 3−1 mod 53 = 18
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·3 (mod 53).
r = 53, t = 0
r = 3, q = 17, t = 1
r = 2, q = 1, t = −17
r = 1, q = 2, t = 18
Inverse
1 ≡ 18·3 (mod 53)
x = 18
check: 3·18 = 54 = 1·53 + 1
Answer
3−1 mod 53 = 18
8.Number theory & modular arithmetic
Answer: 19107 mod 69 = 37
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.