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I have given n, e, c, and 2d+phi(n). any idea how to proceed to solve that RSA. for values click on this:- https://pastebin.com/mdeSdfzD

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    $\begingroup$ What have you tried, where are you stuck? This question shows absolutely no effort. $\endgroup$
    – Maeher
    Feb 17 '18 at 3:40
  • $\begingroup$ I have tried the same thing which is given in below comment. ed = 1+k*phi(n). but there is another variable k, after which I am not able to understand what to do next. $\endgroup$
    – Mark
    Feb 17 '18 at 13:19
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    $\begingroup$ he's cheating on easyctf. easyctf.com $\endgroup$
    – Kevin
    Feb 17 '18 at 21:46
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p = 179625470269984575211291664517722616039787182426068867084101295486932993749787129633737153741496747351573119936946306867775629661781307731728093871384501392452057213703678422306940017327268991666011108647323837661852433288372793079254503318017584711837671484712835436166606958277493346384859583473813750793903

q = 165951900395573838473861177483977329257584913855055564650331190368208599646122670409920222286702764849767512840736219293650470102124983099032047151432801085844142882059163396523057801419384753318562498454796480286799673413582017708953121180295255267080043853111452400359241673145348291220080289048917975182701

Hint: $(2d + \phi(n))\cdot e \equiv 2 \pmod{\phi(n)} \implies (2d + \phi(n))\cdot e - 2$ is a multiple of $\phi(n)$

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    $\begingroup$ How you get p and q, Can you explain more, I am not able to understand it, it's important for me $\endgroup$
    – Mark
    Feb 17 '18 at 13:12
  • $\begingroup$ See also crypto.stackexchange.com/questions/6361 $\endgroup$
    – user94293
    Feb 18 '18 at 22:15
  • $\begingroup$ This doesn't make sense-- his 2d+phi(n) multiplied by e in his example is a multi-hundred digit number. And 2 mod (number > 2) always equals 2. $\endgroup$
    – Aaron Esau
    Feb 19 '18 at 19:03
  • $\begingroup$ @Arin: Above notation is for congruence. It means $[(2d+\phi(n))\cdot e] \bmod \phi(N) = 2 \bmod \phi(N)$. $\endgroup$
    – user94293
    Feb 19 '18 at 19:16
  • $\begingroup$ @Mark And as soon as you know a multiple of $\phi(N)$ you can factor $N$ using the approach as when you know the public and private key but not the primes. $\endgroup$
    – SEJPM
    Feb 21 '18 at 10:03

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