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 ¬p → (q ∨ r) by contradiction. Which statement do you assume at the start? Click it.
¬p → (q ∨ r)
1¬¬p ∧ (q ∨ r)
2¬p ∧ ¬(q ∨ r)
3¬¬p → ¬(q ∨ r)
4¬¬p ∧ ¬(q ∨ r)
2.Propositional logic & proofsHard
Which statement is equivalent to ¬∃x A(x), with the negation moved all the way inside? Click it.
¬∃x A(x)
1∃x A(x)
2∀x A(x)
3∃x ¬A(x)
4∀x ¬A(x)
3.Sets, relations & functionsMedium
A survey of 64 people finds that 34 own a laptop (L), 34 a tablet (T) and 31 a smartwatch (W). 17 own L and T, 14 own L and W, 14 own T and W, and 6 own all three. How many are in L only?
|U| = 64
|L − (T ∪ W)| =
4.Sets, relations & functionsHard
An equivalence relation R on {1, …, 9} has the classes {2, 5, 7}, {1, 4, 6, 8}, {3, 9}. How many ordered pairs does R contain? Call the number |R|.
{2, 5, 7}{1, 4, 6, 8}{3, 9}
|R| =
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
How many closed walks of length 3 start and end at D? A walk may repeat vertices and edges. Call the number N.
M = adjacency matrixL = 3
N =
7.Number theory & modular arithmeticMedium
Solve 15x ≡ 3 (mod 23). Give the solution x with 0 < x < 23.
15x ≡ 3 (mod 23)
x =
8.Number theory & modular arithmeticHard
79 is prime. Use Fermat’s little theorem to find 1868954 mod 79.
p = 79k = 68954
1868954 mod 79 =
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 = ¬p, b = q ∨ r
Negation of the implication
¬(a → b) ≡ a ∧ ¬b = ¬p ∧ ¬(q ∨ r)
Answer
¬p ∧ ¬(q ∨ r)
If the assumption leads to a contradiction, the implication must be true.
2.Propositional logic & proofs
Answer: ∀x ¬A(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 A(x) ≡ ∀x ¬A(x)
Answer
∀x ¬A(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: |L − (T ∪ W)| = 9
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: 6
L ∩ T only: 17 − 6 = 11
L ∩ W only: 14 − 6 = 8
T ∩ W only: 14 − 6 = 8
Single sets
L only: 34 − 11 − 8 − 6 = 9
T only: 34 − 11 − 8 − 6 = 9
W only: 31 − 8 − 8 − 6 = 9
Answer
|L − (T ∪ W)| = 9
Answer
|L − (T ∪ W)| = 9
4.Sets, relations & functions
Answer: |R| = 29
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| = 3² + 4² + 2² = 29
Answer
|R| = 29
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: N = 2
Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.
Adjacency matrix
A: 0 1 1 0 0
B: 1 0 1 1 1
C: 1 1 0 0 0
D: 0 1 0 0 1
E: 0 1 0 1 0
One more step
(M³)DD = 1·0 + 1·1 + 1·0 + 2·0 + 1·1 = 2
Answer
N = 2
7.Number theory & modular arithmetic
Answer: x = 14
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 15
r = 23, t = 0
r = 15, q = 1, t = 1
r = 8, q = 1, t = −1
r = 7, q = 1, t = 2
r = 1, q = 7, t = −3
15−1 ≡ 20 (mod 23)
Solve
x ≡ 20·3 = 60 ≡ 14 (mod 23)
check: 15·14 = 210 ≡ 3 (mod 23)
Answer
x = 14
8.Number theory & modular arithmetic
Answer: 1868954 mod 79 = 8
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.