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 ¬(r ∧ p)? Click it.
¬(r ∧ p)
1¬r ∨ ¬p
2¬r ∧ p
3¬r ∨ p
4¬r ∧ ¬p
2.Propositional logic & proofsHard
Which statement is equivalent to ¬∀x P(x), with the negation moved all the way inside? Click it.
¬∀x P(x)
1∀x ¬P(x)
2∃x P(x)
3∀x P(x)
4∃x ¬P(x)
3.Sets, relations & functionsMedium
Of 44 first-year students, 21 passed Mathematics (M), 15 passed Physics (P) and 30 passed Chemistry (C). 4 passed M and P, 14 passed M and C, 7 passed P and C, and 1 passed all three. How many are in M only?
|U| = 44
|M − (P ∪ C)| =
4.Sets, relations & functionsHard
How many injective (one-to-one) functions f: A → B are there if |A| = 4 and |B| = 6? Call the number N.
|A| = 4|B| = 6
N =
5.Graph theoryMedium
How many different Hamilton circuits H does the complete graph K4 have? Circuits that differ only in the starting vertex or the direction count as the same.
K4
H =
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 62 modulo 67: the number x with 0 < x < 67 and 62·x ≡ 1 (mod 67).
a = 62n = 67
62−1 mod 67 =
8.Number theory & modular arithmeticHard
Encrypt the message m = 140 with the RSA public key (n, e) = (779, 17). Find the ciphertext c.
n = 779e = 17m = 140
c =
3Solutions
1.Propositional logic & proofs
Answer: ¬r ∨ ¬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
De Morgan: ¬(a ∧ b) ≡ ¬a ∨ ¬b. The ∧ turns into ∨ and both parts are negated.
¬(r ∧ p) ≡ ¬r ∨ ¬p
Check with the truth table
p = 1, r = 1: 0 | 0 0 1 0
p = 1, r = 0: 1 | 1 1 1 0
p = 0, r = 1: 1 | 1 0 0 0
p = 0, r = 0: 1 | 1 0 1 1
Answer
¬r ∨ ¬p
The columns after the bar are the options in the order shown.
2.Propositional logic & proofs
Answer: ∃x ¬P(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) ≡ ∃x ¬P(x)
Answer
∃x ¬P(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: |M − (P ∪ C)| = 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|.
Centre and pairs
M ∩ P ∩ C: 1
M ∩ P only: 4 − 1 = 3
M ∩ C only: 14 − 1 = 13
P ∩ C only: 7 − 1 = 6
Single sets
M only: 21 − 3 − 13 − 1 = 4
P only: 15 − 3 − 6 − 1 = 5
C only: 30 − 13 − 6 − 1 = 10
Answer
|M − (P ∪ C)| = 4
Answer
|M − (P ∪ C)| = 4
4.Sets, relations & functions
Answer: N = 360
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.
Count
N = 6!/(6 − 4)! = 360
Answer
N = 360
5.Graph theory
Answer: H = 3
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.
Count
H = (4 − 1)!/2 = 6/2 = 3
Answer
H = 3
6.Graph theory
Answer: d(F) = 18
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: 5
C (5): E: 11, G: 16
B (8): D: 21
E (11): F: 18
G (16)
F (18)
D (21)
Shortest path
A → C → E → F
d(F) = 18
Answer
d(F) = 18
7.Number theory & modular arithmetic
Answer: 62−1 mod 67 = 40
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·62 (mod 67).
r = 67, t = 0
r = 62, q = 1, t = 1
r = 5, q = 12, t = −1
r = 2, q = 2, t = 13
r = 1, q = 2, t = −27
Inverse
1 ≡ −27·62 (mod 67)
x = 40
check: 62·40 = 2480 = 37·67 + 1
Answer
62−1 mod 67 = 40
8.Number theory & modular arithmetic
Answer: c = 334
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.