You Get Extra Information 1

Description

By hellopir2

@quasar0147 solve it in 3 operations now.

Downloads: main.py, output.txt

Solution

I used https://quickmath.com to solve for p,q.

Expressions:

p * q = 83790217241770949930785127822292134633736157973099853931383028198485119939022553589863171712515159590920355561620948287649289302675837892832944404211978967792836179441682795846147312001618564075776280810972021418434978269714364099297666710830717154344277019791039237445921454207967552782769647647208575607201
p + q + q = 26565552874478429895594150715835574472819014534271940714512961970223616824812349678207505829777946867252164956116701692701674023296773659395833735044077013

Answers:

p = 10307593182692464771859444251060107096737374203269378602523841560237645162159897774629661654952776237523790091862810352641745767095280638930754828620479591
q = 8128979845892982561867353232387733688040820165501281055994560204992985831326225951788922087412585314864187432126945670029964128100746510232539453211798711

Youcan grab numbers from image alt text with Inpect Element.

Perform classic RSA decryption:

from Crypto.Util.number import long_to_bytes

n = 83790217241770949930785127822292134633736157973099853931383028198485119939022553589863171712515159590920355561620948287649289302675837892832944404211978967792836179441682795846147312001618564075776280810972021418434978269714364099297666710830717154344277019791039237445921454207967552782769647647208575607201
c = 55170985485931992412061493588380213138061989158987480264288581679930785576529127257790549531229734149688212171710561151529495719876972293968746590202214939126736042529012383384602168155329599794302309463019364103314820346709676184132071708770466649702573831970710420398772142142828226424536566463017178086577
e = 65537

# Step 1: Calculate p And q
p = 10307593182692464771859444251060107096737374203269378602523841560237645162159897774629661654952776237523790091862810352641745767095280638930754828620479591
q = 8128979845892982561867353232387733688040820165501281055994560204992985831326225951788922087412585314864187432126945670029964128100746510232539453211798711

# Step 2: Calculate The Private Key `d`
phi = (p - 1) * (q - 1)
d = pow(e, -1, phi)

# Step 3: Decrypt Ciphertext
m = pow(c, d, n)

# Step 4: Print Flag
print(f"Flag: {long_to_bytes(m).decode()}")

Flag:amateursCTF{harder_than_3_operations?!?!!}