Computer Science Fundamentals: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

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1Formulas and methods

Number systems

Place values
binary 2k, octal 8k, hex 16k
To decimal
Σ dk·bk
From decimal
divide by the base, read the remainders bottom-up
Hex ↔ binary
one hex digit = 4 bits
Two’s complement
invert the bits and add 1

Propositional logic & proofs

Rows of a truth table
2ⁿ
Implication
p → q ≡ ¬p ∨ q ≡ ¬q → ¬p
Negated implication
¬(p → q) ≡ p ∧ ¬q
Biconditional
p ↔ q ≡ (p → q) ∧ (q → p)
De Morgan
¬(p ∧ q) ≡ ¬p ∨ ¬q, ¬(p ∨ q) ≡ ¬p ∧ ¬q
Distributive laws
p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r), p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
Negated quantifiers
¬∀x P(x) ≡ ∃x ¬P(x), ¬∃x P(x) ≡ ∀x ¬P(x)
Modus ponens, modus tollens
p → q, p ⊢ q; p → q, ¬q ⊢ ¬p

Method

  1. Count the different variables: the truth table has 2ⁿ rows, starting with all of them 1.
  2. Give every subformula its own column and fill the columns from the inside out.
  3. Apply the rule of each connective: → is 0 only for 1 → 0, ↔ is 1 when both sides agree.
  4. For equivalences, compare the final columns row by row; for an argument, look only at the rows in which every premise is 1.
  5. To negate a quantified statement, move ¬ inwards, swap ∀ and ∃, and negate the inner formula.

Digital logic: gates, K-maps & flip-flops

Basic gates
A·B, A + B, A′, (A·B)′, (A + B)′, A ⊕ B
Boolean rules
X + X′ = 1, X·X′ = 0, X + XY = X, XY + XY′ = X
De Morgan
(X·Y)′ = X′ + Y′, (X + Y)′ = X′·Y′
Karnaugh map order
00, 01, 11, 10 (Gray code)
Group of 2ᵏ cells
loses k variables
4-to-1 multiplexer
F = A′B′I0 + A′BI1 + AB′I2 + ABI3
Flip-flops
D: Q⁺ = D, T: Q⁺ = T ⊕ Q, JK: Q⁺ = JQ′ + K′Q
Counters
n-bit: 2ⁿ states; mod-N: smallest n with 2ⁿ ≥ N; fout = fclk/N
Ripple counter
fmax = 1/(n·tpd)
Ring and Johnson counters
n and 2n states

Method

  1. For a circuit, work from the inputs to the output one gate at a time and write the value on every wire, or write its expression and fill in the truth table.
  2. For simplification, put the 1s (and the don’t-cares as X) into a Karnaugh map with Gray-code rows and columns: 00, 01, 11, 10.
  3. Cover the 1s with as few and as large groups of 1, 2, 4, 8 or 16 cells as possible, wrapping round the edges; each group is one product term made of the variables that stay constant. Group the 0s for the POS form.
  4. For flip-flops and counters, read the inputs at each active clock edge, write the state after every edge, and remember that an n-bit counter wraps round after 2ⁿ states.

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

  1. Reduce every number modulo n first; remainders lie between 0 and n − 1.
  2. For a gcd, divide repeatedly with Euclid’s algorithm; the last nonzero remainder is the gcd.
  3. For an inverse, run the extended algorithm and keep the coefficient of a; add n if it is negative.
  4. Solve a·x ≡ b by multiplying with a⁻¹; combine two congruences by writing x = r₁ + m₁k.
  5. 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).

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

  1. Write down the degree of every vertex; the degrees add up to twice the number of edges.
  2. For Euler questions, check that the graph is connected and count the vertices with odd degree.
  3. For trees, use n − 1 edges and the handshake lemma.
  4. For Dijkstra, keep a table of tentative distances and settle the smallest unsettled vertex each round.
  5. 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).

IPv4 subnetting

Hosts per subnet
2(32 − prefix) − 2 (/31: 2, /32: 1)
Block size
256 − (mask octet)
Network address
IP AND mask
Broadcast address
IP OR (NOT mask): all host bits set to 1
Subnets from n borrowed bits
2n

2Practice problems

1.Number systemsMedium
8-bit two's complement0−12806403201608040201

Set the bits so they store −35 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −358 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0011 1000 AND 0011 0110. What is the result's value R in decimal?

0011 1000AND0011 0110
R =
3.Propositional logic & proofsMedium
pqr(¬r ∨ p) ∨ q¬((¬r ∨ p) ∨ q)(¬p → (r → q)) → ¬((¬r ∨ p) ∨ q)111?110?101?100?011?010?001?000?

In how many of the 8 rows of its truth table is (¬p → (r → q)) → ¬((¬r ∨ p) ∨ q) true? Call this number N.

(¬p → (r → q)) → ¬((¬r ∨ p) ∨ q)
N =
4.Propositional logic & proofsHard
∃x (¬P(x) ∧ ¬Q(x))∀x (¬P(x) ∨ ¬Q(x))∃x (¬P(x) ∨ ¬Q(x))∀x (¬P(x) ∧ ¬Q(x))

Which statement is equivalent to ¬∀x (P(x) ∨ Q(x)), with the negation moved all the way inside? Click it.

¬∀x (P(x) ∨ Q(x))
  1. 1∃x (¬P(x) ∧ ¬Q(x))
  2. 2∀x (¬P(x) ∨ ¬Q(x))
  3. 3∃x (¬P(x) ∨ ¬Q(x))
  4. 4∀x (¬P(x) ∧ ¬Q(x))
5.Digital logic: gates, K-maps & flip-flopsMedium
MUX0I0​0I1​1I2​1I3​ABS1​S0​F

A 4-to-1 multiplexer has its select inputs S₁ = A and S₀ = B, and its data inputs tied to I₀ = 0, I₁ = 0, I₂ = 1, I₃ = 1. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.

I₀ = 0I₁ = 0I₂ = 1I₃ = 1
N =
6.Digital logic: gates, K-maps & flip-flopsHard
DQ3​Q3​DQ2​Q2​DQ1​Q1​DQ0​Q0​CLKRING

A 4-bit ring counter Q₃Q₂Q₁Q₀ starts at 1001. At every clock edge the bits shift one place to the right (Q₃ → Q₂ → Q₁ → Q₀) and Q₃ receives Q₀. Give the state Q = Q₃Q₂Q₁Q₀ as a decimal number after 9 clock edges.

Q₃Q₂Q₁Q₀ = 10019
Q =
7.Number theory & modular arithmeticMedium

Find the inverse of 52 modulo 73: the number x with 0 < x < 73 and 52·x ≡ 1 (mod 73).

a = 52n = 73
52−1 mod 73 =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 38166 mod 89.

a = 38k = 166n = 89
38166 mod 89 =
9.Algorithms & complexityMedium

What is the time complexity of the standard dynamic programming algorithm for the longest common subsequence of strings of lengths m and n?

  1. AO(2m+n)
  2. BO(mn)
  3. CO(m + n)
  4. DO(mn log n)
10.Algorithms & complexityHard

Which statement about little-o notation is correct?

  1. An² = o(n)
  2. Bn = o(n), since every function is little-o of itself
  3. Cn = o(n²), but n is not o(n)
  4. D2n = o(n), because 2n grows slowly
11.Data structuresMedium

In a circular queue stored in an array of capacity n, how is the next position after index i computed?

  1. A2i + 1
  2. B(i − 1) mod n
  3. Ci + 1
  4. D(i + 1) mod n
12.Data structuresHard

You must keep the k smallest values from a stream of n numbers using O(k) memory. Which approach works in O(n log k) time?

  1. AA sorted array of all n values
  2. BA FIFO queue of the last k values
  3. CA min-heap holding all n values, from which the k smallest are popped at the end
  4. DA max-heap of size k, replacing its root when a smaller value arrives
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K7 have? Circuits that differ only in the starting vertex or the direction count as the same.

K7
H =
14.Graph theoryHard
123121479915ABCDEFG

Use Dijkstra’s algorithm to find the length d(D) of the shortest path from A to D in the weighted graph.

start: Aend: D
d(D) =
15.Databases & SQLMedium

What does the functional dependency A → B mean?

  1. AEach value of A is associated with exactly one value of B
  2. BA is a foreign key referencing B
  3. CA and B are both primary keys
  4. DEach value of B is associated with exactly one value of A
16.Databases & SQLHard

Table employee(emp_id, dept_id, dept_name) has the key emp_id, and dept_id → dept_name. Which statement is correct?

  1. AIt is in 2NF but not in 3NF
  2. BIt is not in 2NF
  3. CIt is in BCNF and therefore in 3NF
  4. DIt is not in 1NF
17.Computer networksMedium

Why is UDP often preferred for live video calls?

  1. ALow delay matters more than retransmitting every lost packet
  2. BUDP is required for all video
  3. CUDP encrypts data automatically
  4. DUDP guarantees every packet arrives, so the picture never freezes
18.Computer networksHard

An 8-port switch (no VLANs) has one host on each port. How many collision domains and broadcast domains are there?

  1. A1 collision domain, 8 broadcast domains
  2. B8 collision domains, 8 broadcast domains
  3. C8 collision domains, 1 broadcast domain
  4. D1 collision domain, 1 broadcast domain
19.IPv4 subnettingMedium
/24 → /2711111111111111111111111111100000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 8 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/248 subnets
H =
20.IPv4 subnettingHard
third octet = 70012816403201608141201networkhost

A host has the address 172.20.70.231/23. What is the third octet x of its subnet's broadcast address?

172.20.70.231/23
x =

3Solutions

1.Number systems

Answer: U = 221

Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.

Invert and add one

35 = 0010 0011
inverted: 1101 1100
+1 → 1101 1101

Read it as unsigned

1101 1101₂ = 256 − 35 = 221

Answer

U = 221
2.Number systems

Answer: R = 48

Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.

Bit by bit

AND gives 1 only where both bits are 1.

0011 1000
AND 0011 0110
= 0011 0000

Value of the result

0011 0000₂ = 48

Answer

R = 48
3.Propositional logic & proofs

Answer: N = 1

Truth tables: a formula with n different variables has 2ⁿ rows, one for every combination of truth values (1 = true, 0 = false). Work from the inside out and give every subformula its own column. ¬a flips the value; a ∧ b is 1 only when both are 1; a ∨ b is 0 only when both are 0; a → b is 0 only when a = 1 and b = 0; a ↔ b is 1 when both sides are equal.

Truth table row by row

p = 1, q = 1, r = 1: ¬p = 0, r → q = 1, ¬p → (r → q) = 1, ¬r = 0, ¬r ∨ p = 1, (¬r ∨ p) ∨ q = 1, ¬((¬r ∨ p) ∨ q) = 0, (¬p → (r → q)) → ¬((¬r ∨ p) ∨ q) = 0
p = 1, q = 1, r = 0: ¬p = 0, r → q = 1, ¬p → (r → q) = 1, ¬r = 1, ¬r ∨ p = 1, (¬r ∨ p) ∨ q = 1, ¬((¬r ∨ p) ∨ q) = 0, (¬p → (r → q)) → ¬((¬r ∨ p) ∨ q) = 0
p = 1, q = 0, r = 1: ¬p = 0, r → q = 0, ¬p → (r → q) = 1, ¬r = 0, ¬r ∨ p = 1, (¬r ∨ p) ∨ q = 1, ¬((¬r ∨ p) ∨ q) = 0, (¬p → (r → q)) → ¬((¬r ∨ p) ∨ q) = 0
p = 1, q = 0, r = 0: ¬p = 0, r → q = 1, ¬p → (r → q) = 1, ¬r = 1, ¬r ∨ p = 1, (¬r ∨ p) ∨ q = 1, ¬((¬r ∨ p) ∨ q) = 0, (¬p → (r → q)) → ¬((¬r ∨ p) ∨ q) = 0
p = 0, q = 1, r = 1: ¬p = 1, r → q = 1, ¬p → (r → q) = 1, ¬r = 0, ¬r ∨ p = 0, (¬r ∨ p) ∨ q = 1, ¬((¬r ∨ p) ∨ q) = 0, (¬p → (r → q)) → ¬((¬r ∨ p) ∨ q) = 0
p = 0, q = 1, r = 0: ¬p = 1, r → q = 1, ¬p → (r → q) = 1, ¬r = 1, ¬r ∨ p = 1, (¬r ∨ p) ∨ q = 1, ¬((¬r ∨ p) ∨ q) = 0, (¬p → (r → q)) → ¬((¬r ∨ p) ∨ q) = 0
p = 0, q = 0, r = 1: ¬p = 1, r → q = 0, ¬p → (r → q) = 0, ¬r = 0, ¬r ∨ p = 0, (¬r ∨ p) ∨ q = 0, ¬((¬r ∨ p) ∨ q) = 1, (¬p → (r → q)) → ¬((¬r ∨ p) ∨ q) = 1
p = 0, q = 0, r = 0: ¬p = 1, r → q = 1, ¬p → (r → q) = 1, ¬r = 1, ¬r ∨ p = 1, (¬r ∨ p) ∨ q = 1, ¬((¬r ∨ p) ∨ q) = 0, (¬p → (r → q)) → ¬((¬r ∨ p) ∨ q) = 0

Count the rows with value 1

(¬p → (r → q)) → ¬((¬r ∨ p) ∨ q): 0, 0, 0, 0, 0, 0, 1, 0
N = 1

Answer

N = 1
4.Propositional logic & proofs

Answer: ∃x (¬P(x) ∧ ¬Q(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) ∨ Q(x)) ≡ ∃x (¬P(x) ∧ ¬Q(x))

Answer

∃x (¬P(x) ∧ ¬Q(x))

The other options either keep a quantifier unchanged, change the order of the quantifiers, or negate the inner formula wrongly.

5.Digital logic: gates, K-maps & flip-flops

Answer: N = 4

Multiplexer: a 2ⁿ-to-1 multiplexer passes one of its data inputs to the output; the select lines, read as a binary number, pick which one. For a 4-to-1 MUX with selects S₁S₀ = AB, F = A′B′·I₀ + A′B·I₁ + AB′·I₂ + AB·I₃. Tying the data inputs to 0, 1, C or C′ makes the MUX produce any function of A, B and C, which is a common way to build logic with one chip.

Output expression

F = AB′ + AB

Rows with F = 1

ABC = 100, 101, 110, 111
N = 4

Answer

N = 4
6.Digital logic: gates, K-maps & flip-flops

Answer: Q = 12

Flip-flops and counters: a flip-flop changes only at the active clock edge. D copies its input (Q⁺ = D); T toggles when T = 1; JK holds for 00, resets for 01, sets for 10 and toggles for 11. An n-bit binary counter has 2ⁿ states and counts modulo 2ⁿ; a mod-N counter goes 0, 1, …, N − 1 and back to 0, and needs the smallest n with 2ⁿ ≥ N flip-flops. Each stage of a ripple counter halves the frequency, so the last output of a mod-N counter runs at fclk/N. A Johnson counter of n flip-flops has 2n states, a ring counter n.

State after every edge

1001 → 1100 → 0110 → 0011 → 1001 → 1100 → 0110 → 0011 → 1001 → 1100

Decimal value

1100₂ = 12

Answer

Q = 12
7.Number theory & modular arithmetic

Answer: 52−1 mod 73 = 66

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·52 (mod 73).

r = 73, t = 0
r = 52, q = 1, t = 1
r = 21, q = 2, t = −1
r = 10, q = 2, t = 3
r = 1, q = 10, t = −7

Inverse

1 ≡ −7·52 (mod 73)
x = 66
check: 52·66 = 3432 = 47·73 + 1

Answer

52−1 mod 73 = 66
8.Number theory & modular arithmetic

Answer: 38166 mod 89 = 18

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.

Binary exponent

166 = 101001102 = 128 + 32 + 4 + 2

Repeated squaring

381 ≡ 38 (mod 89)
382 ≡ 20 (mod 89)
384 ≡ 44 (mod 89)
388 ≡ 67 (mod 89)
3816 ≡ 39 (mod 89)
3832 ≡ 8 (mod 89)
3864 ≡ 64 (mod 89)
38128 ≡ 2 (mod 89)

Multiply the needed powers

38166 ≡ 20·44·8·2 ≡ 18 (mod 89)

Answer

38166 mod 89 = 18
9.Algorithms & complexity

Answer: B · O(mn)

Why this is the answer

The table has (m + 1)(n + 1) cells and each is computed in O(1) from its neighbours.

Answer

B · O(mn)
10.Algorithms & complexity

Answer: C · n = o(n²), but n is not o(n)

Why this is the answer

f = o(g) means f(n) / g(n) tends to 0; n / n² tends to 0, but n / n = 1 does not.

Answer

C · n = o(n²), but n is not o(n)
11.Data structures

Answer: D · (i + 1) mod n

Why this is the answer

The modulo operation wraps index n − 1 back to 0, so the array is reused without shifting elements.

Answer

D · (i + 1) mod n
12.Data structures

Answer: D · A max-heap of size k, replacing its root when a smaller value arrives

Why this is the answer

The max-heap's root is the largest kept value; any smaller new value replaces it in O(log k), so the heap always holds the k smallest so far.

Answer

D · A max-heap of size k, replacing its root when a smaller value arrives
13.Graph theory

Answer: H = 360

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 = (7 − 1)!/2 = 720/2 = 360

Answer

H = 360
14.Graph theory

Answer: d(D) = 13

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: 12, C: 3
C (3): E: 5, G: 12
E (5): D: 19
B (12): D: 13
G (12)
D (13): F: 20
F (20)

Shortest path

A → B → D
d(D) = 13

Answer

d(D) = 13
15.Databases & SQL

Answer: A · Each value of A is associated with exactly one value of B

Why this is the answer

A → B means any two rows with the same A value must have the same B value; A determines B.

Answer

A · Each value of A is associated with exactly one value of B
16.Databases & SQL

Answer: A · It is in 2NF but not in 3NF

Why this is the answer

The key has one column, so 2NF holds, but emp_id → dept_id → dept_name is a transitive dependency, which violates 3NF.

Answer

A · It is in 2NF but not in 3NF
17.Computer networks

Answer: A · Low delay matters more than retransmitting every lost packet

Why this is the answer

Retransmitting lost packets would add delay; real-time media tolerates some loss, so UDP-based protocols like RTP are used.

Answer

A · Low delay matters more than retransmitting every lost packet
18.Computer networks

Answer: C · 8 collision domains, 1 broadcast domain

Why this is the answer

Each switch port is its own collision domain, but broadcasts are forwarded to all ports, so the whole switch is one broadcast domain. A hub would give one of each.

Answer

C · 8 collision domains, 1 broadcast domain
19.IPv4 subnetting

Answer: H = 30

IPv4 subnetting: an address has 32 bits. The prefix /p says how many leading bits identify the network; the remaining 32 − p bits number the hosts. A subnet has 2³²⁻ᵖ addresses, but the first (network address) and last (broadcast address) cannot be given to hosts, so 2³²⁻ᵖ − 2 hosts fit. Exceptions: a /31 (point-to-point link, RFC 3021) uses both of its 2 addresses for hosts, and a /32 is one single address (a single host or a loopback).

Bits to borrow

23 = 8 ≥ 8 → b = 3, /27

Hosts per subnet

25 − 2 = 30

Answer

H = 30
20.IPv4 subnetting

Answer: x = 71

IPv4 subnetting: an address has 32 bits. The prefix /p says how many leading bits identify the network; the remaining 32 − p bits number the hosts. A subnet has 2³²⁻ᵖ addresses, but the first (network address) and last (broadcast address) cannot be given to hosts, so 2³²⁻ᵖ − 2 hosts fit. Exceptions: a /31 (point-to-point link, RFC 3021) uses both of its 2 addresses for hosts, and a /32 is one single address (a single host or a loopback).

Block size in the third octet

21 = 2

Subnet of this host

network: 172.20.70.0
broadcast: 172.20.71.255

Answer

x = 71

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