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 −4 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −48 bit
U =
2.Number systemsHard
24 bit1 px1920 px1080 pxM = ? KiB

An uncompressed image is 1920 × 1080 pixels with 24 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

1920 × 1080 px24 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
pqr¬p ∧ r¬(¬p ∧ r)((q ∧ r) ∨ ¬p) → ¬(¬p ∧ r)111?110?101?100?011?010?001?000?

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

((q ∧ r) ∨ ¬p) → ¬(¬p ∧ r)
N =
4.Propositional logic & proofsHard
pr¬pr ∧ q

The premises r ∨ q; r → p; q → p are given. Which conclusion follows from them? Click it.

r ∨ qr → pq → p
  1. 1p
  2. 2r
  3. 3¬p
  4. 4r ∧ q
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC000111100101324576

Simplify F = A′B′C + ABC + AC′ to its minimal sum-of-products form. How many literals L does the simplified expression contain?

F = A′B′C + ABC + AC′
L =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKD12345678

A positive-edge-triggered D flip-flop starts with Q = 0 and receives the inputs shown at 8 rising clock edges. How many times N does Q change state?

Q₀ = 0D
N =
7.Number theory & modular arithmeticMedium

Solve 3x ≡ 7 (mod 23). Give the solution x with 0 < x < 23.

3x ≡ 7 (mod 23)
x =
8.Number theory & modular arithmeticHard
mc = me​ mod n374m = cd​ mod n?(n, e) = (403, 7)d

An RSA key uses p = 13, q = 31 and e = 7. Decrypt the ciphertext c = 374: find the message m.

p = 13q = 31e = 7c = 374
m =
9.Algorithms & complexityMedium

Why is the base of a logarithm usually left out in Big O, as in O(log n)?

  1. ALogarithms with different bases differ only by a constant factor
  2. BThe base only matters for small n
  3. CAll logarithms in computer science have base 10, so it is implied
  4. DLogarithms are always rounded to integers
10.Algorithms & complexityHard

Which of these satisfiability problems is known to be solvable in polynomial time?

  1. AGeneral SAT on formulas in conjunctive normal form
  2. BCircuit satisfiability
  3. C3-SAT, where each clause has exactly three literals
  4. D2-SAT, where each clause has at most two literals
11.Data structuresMedium

In a binary heap stored in an array starting at index 0, where is the parent of the node at index i?

  1. AAt index ⌊(i − 1) / 2⌋
  2. BAt index i − 1
  3. CAt index 2i + 1
  4. DAt index ⌊i / 2⌋ + 1
12.Data structuresHard

Kahn's algorithm for topological sorting repeatedly removes vertices with in-degree 0. Which structure holds those vertices?

  1. AAn adjacency matrix recording which vertices have already been removed
  2. BA max-heap ordered by out-degree
  3. CA queue (or any collection) of vertices whose in-degree is 0
  4. DA stack of all visited edges
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K5 have? Circuits that differ only in the starting vertex or the direction count as the same.

K5
H =
14.Graph theoryHard
4111111213134158ABCDEFG

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

Table a has 3 rows and table b has 4 rows. How many rows does SELECT * FROM a CROSS JOIN b return?

  1. A3
  2. B7
  3. C12
  4. D4
16.Databases & SQLHard

x is NULL and y = 1. What is the value of the condition (x > 5 OR y = 1)?

  1. AFALSE
  2. BUNKNOWN
  3. CNULL, so the row is always rejected
  4. DTRUE
17.Computer networksMedium

What does ARP do?

  1. ARoutes packets on the internet
  2. BFinds the MAC address that belongs to an IP address on the local network
  3. CAssigns IP addresses automatically to hosts when they join the local network
  4. DResolves domain names
18.Computer networksHard

Head-of-line blocking in HTTP/2 over TCP occurs because:

  1. AServers cannot handle parallel requests
  2. BHTTP/2 uses UDP
  3. CA lost TCP segment delays all multiplexed streams until it is retransmitted
  4. DEach request needs its own new TCP connection, so later requests wait in a queue
19.IPv4 subnettingMedium
/24 → /2811111111111111111111111111110000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 10 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/2410 subnets
H =
20.IPv4 subnettingHard

The network 192.168.81.0/24 is split with VLSM for these LANs: A: 73 hosts, B: 13 hosts, C: 19 hosts, D: 35 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN C?

A: 73B: 13C: 19D: 35
x =

3Solutions

1.Number systems

Answer: U = 252

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

4 = 0000 0100
inverted: 1111 1011
+1 → 1111 1100

Read it as unsigned

1111 1100₂ = 256 − 4 = 252

Answer

U = 252
2.Number systems

Answer: M = 6075 KiB

Positional number systems: each digit is worth its value times a power of the base. In binary the places are worth 1, 2, 4, 8, 16, …; in hexadecimal 1, 16, 256, …, with A–F standing for 10–15. In 8-bit two's complement the leftmost bit is worth −128 instead of +128, which is how negative numbers are stored.

Bits and bytes

1920·1080·24 = 49766400 bit = 6220800 B

Convert to KiB

6220800/1024 = 6075 KiB

Answer

M = 6075 KiB
3.Propositional logic & proofs

Answer: N = 6

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

Count the rows with value 1

((q ∧ r) ∨ ¬p) → ¬(¬p ∧ r): 1, 1, 1, 1, 0, 1, 0, 1
N = 6

Answer

N = 6
4.Propositional logic & proofs

Answer: p

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

p = 1, q = 1, r = 1: p = 1, r = 1, ¬p = 0, r ∧ q = 1
p = 1, q = 1, r = 0: p = 1, r = 0, ¬p = 0, r ∧ q = 0
p = 1, q = 0, r = 1: p = 1, r = 1, ¬p = 0, r ∧ q = 0

Conclusion

p (proof by cases)

Answer

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

Answer: L = 7

Boolean algebra: the rules that simplify expressions are X + X′ = 1 and X·X′ = 0, X + X = X, absorption X + X·Y = X, the combining rule X·Y + X·Y′ = X, X + X′Y = X + Y and the consensus theorem XY + X′Z + YZ = XY + X′Z. De Morgan's laws turn (X·Y)′ into X′ + Y′ and (X + Y)′ into X′·Y′. A Karnaugh map applies these rules graphically: every group of two neighbouring cells is one use of XY + XY′ = X.

Minterms of the expression

A′B′C → m(1)
ABC → m(7)
AC′ → m(4, 6)
ABC00011110010011030214051716

Group and read off

F = AC′ + AB + A′B′C

Count the literals

L = 7

Answer

L = 7
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 5

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.

Q after every edge

A D flip-flop copies D at each rising edge: Q⁺ = D.

1: D = 1 → Q = 1
2: D = 0 → Q = 0
3: D = 0 → Q = 0
4: D = 1 → Q = 1
5: D = 1 → Q = 1
6: D = 0 → Q = 0
7: D = 0 → Q = 0
8: D = 1 → Q = 1

Count the changes

0 → 1 → 0 → 0 → 1 → 1 → 0 → 0 → 1
N = 5

Answer

N = 5
7.Number theory & modular arithmetic

Answer: x = 10

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 3

r = 23, t = 0
r = 3, q = 7, t = 1
r = 2, q = 1, t = −7
r = 1, q = 2, t = 8
3−1 ≡ 8 (mod 23)

Solve

x ≡ 8·7 = 56 ≡ 10 (mod 23)
check: 3·10 = 30 ≡ 7 (mod 23)

Answer

x = 10
8.Number theory & modular arithmetic

Answer: m = 101

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.

Private exponent

n = 403, φ(n) = 360
d = 7−1 mod 360 = 103

Decrypt by repeated squaring

103 = 64 + 32 + 4 + 2 + 1
3741 ≡ 374 (mod 403)
3742 ≡ 35 (mod 403)
3744 ≡ 16 (mod 403)
3748 ≡ 256 (mod 403)
37416 ≡ 250 (mod 403)
37432 ≡ 35 (mod 403)
37464 ≡ 16 (mod 403)
374103 ≡ 374·35·16·35·16 ≡ 101 (mod 403)

Message

m = 101
check: 1017 mod 403 = 374

Answer

m = 101
9.Algorithms & complexity

Answer: A · Logarithms with different bases differ only by a constant factor

Why this is the answer

Changing the base multiplies the logarithm by a constant, and Big O ignores constant factors.

Answer

A · Logarithms with different bases differ only by a constant factor
10.Algorithms & complexity

Answer: D · 2-SAT, where each clause has at most two literals

Why this is the answer

2-SAT is in P (linear time via strongly connected components), while 3-SAT is NP-complete.

Answer

D · 2-SAT, where each clause has at most two literals
11.Data structures

Answer: A · At index ⌊(i − 1) / 2⌋

Why this is the answer

Children of i are at 2i + 1 and 2i + 2, so the parent of i is ⌊(i − 1) / 2⌋; with 1-based indexing it is ⌊i / 2⌋.

Answer

A · At index ⌊(i − 1) / 2⌋
12.Data structures

Answer: C · A queue (or any collection) of vertices whose in-degree is 0

Why this is the answer

Ready vertices are kept in a queue; removing one lowers its neighbours' in-degrees, and any that reach 0 are added, giving O(V + E) time.

Answer

C · A queue (or any collection) of vertices whose in-degree is 0
13.Graph theory

Answer: H = 12

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 = (5 − 1)!/2 = 24/2 = 12

Answer

H = 12
14.Graph theory

Answer: d(D) = 5

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: 4
B (4): C: 15, D: 5, G: 8
D (5): E: 17, F: 18
G (8)
C (15)
E (17)
F (18)

Shortest path

A → B → D
d(D) = 5

Answer

d(D) = 5
15.Databases & SQL

Answer: C · 12

Why this is the answer

A cross join gives the Cartesian product, 3 × 4 = 12 rows.

Answer

C · 12
16.Databases & SQL

Answer: D · TRUE

Why this is the answer

x > 5 is UNKNOWN, but OR with TRUE is TRUE regardless, so the row is kept; UNKNOWN AND FALSE would likewise be FALSE.

Answer

D · TRUE
17.Computer networks

Answer: B · Finds the MAC address that belongs to an IP address on the local network

Why this is the answer

ARP broadcasts "who has this IP?" and the owner replies with its MAC address, so frames can be delivered locally.

Answer

B · Finds the MAC address that belongs to an IP address on the local network
18.Computer networks

Answer: C · A lost TCP segment delays all multiplexed streams until it is retransmitted

Why this is the answer

TCP delivers bytes in order, so one loss stalls every stream on the connection. QUIC (HTTP/3) handles streams independently over UDP.

Answer

C · A lost TCP segment delays all multiplexed streams until it is retransmitted
19.IPv4 subnetting

Answer: H = 14

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

24 = 16 ≥ 10 → b = 4, /28

Hosts per subnet

24 − 2 = 14

Answer

H = 14
20.IPv4 subnetting

Answer: x = 192

VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.

Block size per LAN

A: 73 hosts → 27 = 128 (/25)
D: 35 hosts → 26 = 64 (/26)
C: 19 hosts → 25 = 32 (/27)
B: 13 hosts → 24 = 16 (/28)

Allocation, largest first

A: .0 – .127
D: .128 – .191
C: .192 – .223
B: .224 – .239

Answer

x = 192

LAN C: 192.168.81.192/27.

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