CTF crypto challenge playbook — read the provided encryption script like a spec, attack RSA parameter weaknesses, XOR/multi-time-pad oracles, and homemade ciphers with math tooling (z3, sympy, sage). Load when the handout is .py/.sage math, n/e/c values, or an "encryption service" on nc. Signals: "crypto" category, chall.py with Crypto.Util.number, e=3 or e=65537, "encrypted_flag", XOR with a key, nc service that encrypts your input.
Scanned 10/5/2026
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---
name: ctf-crypto
description: >
CTF crypto challenge playbook — read the provided encryption script like a spec, attack RSA
parameter weaknesses, XOR/multi-time-pad oracles, and homemade ciphers with math tooling
(z3, sympy, sage). Load when the handout is .py/.sage math, n/e/c values, or an "encryption
service" on nc. Signals: "crypto" category, chall.py with Crypto.Util.number, e=3 or e=65537,
"encrypted_flag", XOR with a key, nc service that encrypts your input.
domain: ctf
type: technique
stability: learning
modes: [pentest, bugbounty]
severity: medium
cwe: [CWE-327, CWE-780]
tools: [python-pycryptodome, z3, sympy, sage, rsactftool, factordb, xortool, cyberchef]
schema_version: 1
---
# CTF crypto challenges
## When it applies
The handout is a short encryption script (`chall.py`, rarely sage), a set of numbers
(`n, e, c` / intercepted ciphertexts), or an `nc` service that encrypts/signs things for you.
This skill covers the CTF shapes — reading the script for the flaw, XOR games, homemade ciphers,
and math tooling. For the deep attack mechanics it defers to `crypto-rsa-attacks` and
`crypto-oracle-attacks`; run the RSA battery there first.
## Why it works
CTF crypto is a *code review against math*: the script is short, self-contained, and contains
exactly one mistake the author planted — a reused nonce, a factorable `n`, an invertible
homegrown round function, or an oracle the service hands you. Find the one line that looks
"clever" and the challenge is usually done; the rest is algebra.
## Method
1. **Read the script as a spec, top to bottom.** Note exactly what's given (printed values) and
what's hidden (keys, seeds, flag). The flaw is almost always in: key/nonce generation
(`random` instead of `secrets`, timestamp seeds, `getPrime` with small bits, reused `k`),
the composition (same key twice, encrypt-then-leak-something), or the padding (raw RSA, ASCII
armor that leaks length).
2. **Triage encodings before crypto.** Half of "crypto" challenges are encoding stacks: base64/32/
85, hex, morse, brainfuck, base-N with a custom alphabet, repeated base64. CyberChef's magic
wand or a quick decode loop settles these in minutes — don't reach for sage on a base85.
3. **RSA → run the known battery.** Extract `n, e, c`; hand the key to RsaCtfTool + factordb,
then check the CTF-specific shapes explicitly (`crypto-rsa-attacks` for mechanics):
- Multiple keys/ciphertexts in one file → `gcd` pairwise (shared prime), common modulus
(same `n`, two `e`), or Håstad broadcast (same message, small `e`, several `n`).
- `p` and `q` generated from a weak PRNG / shared high bits / `q = next_prime(p)` → Fermat or
regenerate the PRNG stream.
- Extra printed values (`d mod (p-1)`, `p+q`, `dp`, partial bits of `p`) → Coppersmith-style
recovery (sage) or direct algebra; these "hint leaks" are the intended path.
4. **XOR challenges.**
- Single-byte XOR: brute all 256 keys, score by English frequency (`xortool` automates).
- Repeating-key XOR: guess key length by Hamming distance / index of coincidence, solve each
column as single-byte.
- **Multi-time pad** (same keystream XORs several messages, classic `otp` reuse): crib-drag —
XOR two ciphertexts and guess words ("the ", "flag{") at each position; each confirmed crib
extends the keystream.
- XOR with a known plaintext anywhere (a known header, the flag format `flag{`) recovers
keystream bytes directly.
5. **Encryption services are oracles — script them.** `nc` service that encrypts your input:
- Your input prepended to the flag, ECB → byte-at-a-time flag recovery (`crypto-oracle-attacks`).
- Same plaintext always → same ciphertext: codebook lookup / chosen-plaintext mapping.
- Decrypt-anything-but-the-flag → malleability games: bit-flip CBC, RSA blinding
(`c·r^e mod n`, unblind the result).
- It always helps to send structured probes: all-zero blocks, single-byte deltas, length sweeps
(leaks block size and mode).
6. **Homegrown ciphers → model, don't guess.** Custom rounds (add-rotate-xor, matrix mods,
LFSRs, "Feistel-ish"): symbolically execute or invert them.
- Linear operations over GF(2) → write as linear equations, solve with z3 or sage.
- Small state (LFSR) → Berlekamp–Massey from a known keystream.
- S-box + permutation networks → differential or just brute the tiny key space the author used.
- Ask: *is every step invertible given the key? Then the only question is the key space* —
brute 16/24/32-bit keys before doing anything clever.
7. **Bring the math tools.** z3 for constraint-shaped problems (flag bytes satisfying check
equations); sympy for modular algebra and small discrete logs; sage for lattices (LLL /
Coppersmith / hidden-number problems); `long_to_bytes`/`bytes_to_long` endianness discipline
throughout.
## Gotchas
- **The printed "leak" is intentional.** If the script hands you `hint = p ^ q` or
`leak = d & ((1<<512)-1)`, that's the solve path — stop looking for a second bug.
- **Python `random` is not crypto** — `random.seed(time.time())` or Mersenne Twister output
visible anywhere → clone the state (624 consecutive 32-bit outputs fully recover MT19937).
- **`e` and `phi` mistakes** — wrong totient (using `n` for `(p-1)(q-1)`) makes a key that looks
right but fails; verify by decrypting a known value (`crypto-rsa-attacks`).
- **Multi-prime RSA** (`n = pqr`) — factorization tools still work; remember `phi = (p-1)(q-1)(r-1)`.
- **Encoding is not encryption** — if "ciphertext" decodes cleanly to printable text, keep
peeling encodings before assuming a cipher.
- **Byte-vs-int conversion bugs in *your* solver** — leading zero bytes vanish in
`bytes_to_long`; pad the recovered plaintext back to the expected length.
## Verify success
The recovered plaintext contains the flag in the event's format, and you can re-derive it
deterministically (script reproduces the key/plaintext from the given artifacts, not from luck).
## References
Mechanics: `crypto-rsa-attacks` (factordb, Fermat, Wiener, Håstad), `crypto-oracle-attacks`
(padding oracle, ECB byte-at-a-time, length extension); tooling: z3, sage (LLL/Coppersmith),
xortool, CyberChef. Triage via `ctf-methodology`.
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