[HDCTF2019]basic rsa

題目

import gmpy2 from Crypto.Util.number import * from binascii import a2b_hex,b2a_hexflag = "*****************"p = 262248800182277040650192055439906580479 q = 262854994239322828547925595487519915551e = 65533 n = p*qc = pow(int(b2a_hex(flag),16),e,n)print c# 27565231154623519221597938803435789010285480123476977081867877272451638645710

解題
先嘗試將下面的數復制給flag，運行，顯示：

TypeError: a bytes-like object is required, not ‘str’

解決：str 通過 encode() 函數編碼為 bytes

運行得到：7733094235673990295993988034122557233478423989064087878609357727899688710412

有點長，還是嘗試一下
結果意料之中的不對

已經有n、p、q、e了，那按之前的方法解題：

import gmpy2 from Crypto.Util.number import long_to_bytes p = 262248800182277040650192055439906580479 q = 262854994239322828547925595487519915551 e = 65533 c = 27565231154623519221597938803435789010285480123476977081867877272451638645710 n = q*p phi = (q-1) * (p-1) d = gmpy2.invert(e,phi) m = gmpy2.powmod(c,d,n)print(long_to_bytes(m))

答案
flag{B4by_Rs4}

[GUET-CTF2019]BabyRSA

題目
給后綴名打開

p+q : 0x1232fecb92adead91613e7d9ae5e36fe6bb765317d6ed38ad890b4073539a6231a6620584cea5730b5af83a3e80cf30141282c97be4400e33307573af6b25e2ea (p+1)(q+1) : 0x5248becef1d925d45705a7302700d6a0ffe5877fddf9451a9c1181c4d82365806085fd86fbaab08b6fc66a967b2566d743c626547203b34ea3fdb1bc06dd3bb765fd8b919e3bd2cb15bc175c9498f9d9a0e216c2dde64d81255fa4c05a1ee619fc1fc505285a239e7bc655ec6605d9693078b800ee80931a7a0c84f33c851740 e : 0xe6b1bee47bd63f615c7d0a43c529d219 d : 0x2dde7fbaed477f6d62838d55b0d0964868cf6efb2c282a5f13e6008ce7317a24cb57aec49ef0d738919f47cdcd9677cd52ac2293ec5938aa198f962678b5cd0da344453f521a69b2ac03647cdd8339f4e38cec452d54e60698833d67f9315c02ddaa4c79ebaa902c605d7bda32ce970541b2d9a17d62b52df813b2fb0c5ab1a5 enc_flag : 0x50ae00623211ba6089ddfae21e204ab616f6c9d294e913550af3d66e85d0c0693ed53ed55c46d8cca1d7c2ad44839030df26b70f22a8567171a759b76fe5f07b3c5a6ec89117ed0a36c0950956b9cde880c575737f779143f921d745ac3bb0e379c05d9a3cc6bf0bea8aa91e4d5e752c7eb46b2e023edbc07d24a7c460a34a9a

解題
已知p+q、(p+1)(q+1)、e、d、c
可以通過數學計算得到p、q

import gmpy2 from Crypto.Util.number import long_to_bytes#設a=p+q;b=(p+1)*(q-1) a = 0x1232fecb92adead91613e7d9ae5e36fe6bb765317d6ed38ad890b4073539a6231a6620584cea5730b5af83a3e80cf30141282c97be4400e33307573af6b25e2ea #p+q b = 0x5248becef1d925d45705a7302700d6a0ffe5877fddf9451a9c1181c4d82365806085fd86fbaab08b6fc66a967b2566d743c626547203b34ea3fdb1bc06dd3bb765fd8b919e3bd2cb15bc175c9498f9d9a0e216c2dde64d81255fa4c05a1ee619fc1fc505285a239e7bc655ec6605d9693078b800ee80931a7a0c84f33c851740 #(p+1) * (q+1) e = 0xe6b1bee47bd63f615c7d0a43c529d219 d = 0x2dde7fbaed477f6d62838d55b0d0964868cf6efb2c282a5f13e6008ce7317a24cb57aec49ef0d738919f47cdcd9677cd52ac2293ec5938aa198f962678b5cd0da344453f521a69b2ac03647cdd8339f4e38cec452d54e60698833d67f9315c02ddaa4c79ebaa902c605d7bda32ce970541b2d9a17d62b52df813b2fb0c5ab1a5 c = 0x50ae00623211ba6089ddfae21e204ab616f6c9d294e913550af3d66e85d0c0693ed53ed55c46d8cca1d7c2ad44839030df26b70f22a8567171a759b76fe5f07b3c5a6ec89117ed0a36c0950956b9cde880c575737f779143f921d745ac3bb0e379c05d9a3cc6bf0bea8aa91e4d5e752c7eb46b2e023edbc07d24a7c460a34a9a#求p、q q = (a + gmpy2.iroot(a ** 2 - 4 * (b-a-1),2)[0]) // 2 p = a - q

然后就可以進行常規rsa解密了

n = q*p phi = (p-1) * (q-1)d = gmpy2.invert(e,phi) m = gmpy2.powmod(c,d,n) print(long_to_bytes(m))

答案
flag{cc7490e-78ab-11e9-b422-8ba97e5da1fd}

rsa2

題目

N = 101991809777553253470276751399264740131157682329252673501792154507006158434432009141995367241962525705950046253400188884658262496534706438791515071885860897552736656899566915731297225817250639873643376310103992170646906557242832893914902053581087502512787303322747780420210884852166586717636559058152544979471 e = 46731919563265721307105180410302518676676135509737992912625092976849075262192092549323082367518264378630543338219025744820916471913696072050291990620486581719410354385121760761374229374847695148230596005409978383369740305816082770283909611956355972181848077519920922059268376958811713365106925235218265173085import hashlib flag = "flag{" + hashlib.md5(hex(d)).hexdigest() + "}"

解題
需求未知數d

phi = (q-1) * (p-1) d = gmpy2.invert(e,phi)

則，還需將n分解得到p、q
分解得：

p= 9046853915223503351787031888977627106934564043204783593118678181991596316582877057556463152579621699010610569526573031954779520781448550677767565207407183 q=11273732364123571293429600400343309403733952146912318879993851141423284675797325272321856863528776914709992821287788339848962916204774010644058033316303937

運行，出現錯誤
TypeError: Unicode-objects must be encoded before hashing

改正：將hex(d)后加.encode()

輸出flag
flag{8159e6c4abdd3b94ce461ed9a1a24017}

提交，顯示錯誤

檢查一下，d是對的

encode() 方法為字符串類型（str）提供的方法，用于將 str 類型轉換成 bytes 類型，這個過程也稱為“編碼”。

hex() 函數用于將10進制整數轉換成16進制，以字符串形式表示。

MD5簽名是一個哈希函數，把任意長度的數據轉換為一個長度固定的數據串（通常用16進制的字符串表示）。

求出d之后用在線工具解

答案
flag{47bf28da384590448e0b0d23909a25a4}

總結

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