title: For RSA encryption, finding d given p,q, and e?
body: <p><strong>How would I find <code>d</code> given <code>p = 7, q = 11, e = 3</code>?</strong></p>
  
  <p>There are many similar questions to this, but they ask you to calculate your own e. This would be a trivial solution if <code>e = 11</code> or <code>e = 17</code>, but I'm unsure of how to proceed with <code>e = 3</code> specifically.</p>
  
  <p>My understanding goes:</p>
  
  <p>d = e<sup>-1</sup>mod(p-1)(q-1)</p>
  
  <p>d = 3<sup>-1</sup>mod(6)(10)</p>
  
  <p>d * 3 == 1 mod 60</p>
  
  <p>However this leads to 3 not being invertible mod 60, as 3 is not relatively prime to 60. Is there a solution given this value of e=3? Thank you for any help.</p>