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¬(r ∧ p) → ¬¬q
2¬¬q → ¬(r ∧ p)
3¬q ∧ ¬(r ∧ p)
4(r ∧ p) → ¬q
2.Propositional logic & proofsHard
The premises (c ∧ a) → b; c; ¬b are given. Which conclusion follows from them? Click it.
(c ∧ a) → bc¬b
1¬a
2a
3a ∧ ¬b
4b
3.Sets, relations & functionsMedium
Of 39 first-year students, 15 passed Mathematics (M), 18 passed Physics (P) and 21 passed Chemistry (C). 3 passed M and P, 10 passed M and C, 8 passed P and C, and 1 passed all three. How many are in exactly one of the three sets? Call the number n.
|U| = 39
n =
4.Sets, relations & functionsHard
The functions f and g on {1, 2, 3, 4, 5} are given by their tables. Find (f ∘ f)(2).
Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.
|V| = 5|E| = 6
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 72 modulo 89: the number x with 0 < x < 89 and 72·x ≡ 1 (mod 89).
a = 72n = 89
72−1 mod 89 =
8.Number theory & modular arithmeticHard
An RSA key uses the primes p = 11 and q = 23 and the public exponent e = 7. Find the private exponent d (0 < d < φ(n)).
p = 11q = 23e = 7
d =
3Solutions
1.Propositional logic & proofs
Answer: ¬(r ∧ p) → ¬¬q
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, b = r ∧ p
Contrapositive
¬b → ¬a = ¬(r ∧ p) → ¬¬q
Answer
¬(r ∧ p) → ¬¬q
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, b = 0, c = 1: ¬a = 1, a = 0, a ∧ ¬b = 0, b = 0
Conclusion
¬a (modus tollens with a conjunction)
Answer
¬a
3.Sets, relations & functions
Answer: n = 15
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: 1
M ∩ P only: 3 − 1 = 2
M ∩ C only: 10 − 1 = 9
P ∩ C only: 8 − 1 = 7
Single sets
M only: 15 − 2 − 9 − 1 = 3
P only: 18 − 2 − 7 − 1 = 8
C only: 21 − 9 − 7 − 1 = 4
Answer
3 + 8 + 4 = 15
Answer
n = 15
4.Sets, relations & functions
Answer: (f ∘ f)(2) = 1
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.
Inner function
f(2) = 5
Outer function
f(5) = 1
(f ∘ f)(2) = 1
Answer
(f ∘ f)(2) = 1
5.Graph theory
Answer: Euler 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.
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: 1, C: 13
B (1): C: 9, D: 8, E: 12
D (8): F: 10
C (9)
F (10)
E (12)
Shortest path
A → B → D → F
d(F) = 10
Answer
d(F) = 10
7.Number theory & modular arithmetic
Answer: 72−1 mod 89 = 68
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·72 (mod 89).
r = 89, t = 0
r = 72, q = 1, t = 1
r = 17, q = 4, t = −1
r = 4, q = 4, t = 5
r = 1, q = 4, t = −21
Inverse
1 ≡ −21·72 (mod 89)
x = 68
check: 72·68 = 4896 = 55·89 + 1
Answer
72−1 mod 89 = 68
8.Number theory & modular arithmetic
Answer: d = 63
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.