diff --git a/SUMMARY.md b/SUMMARY.md
index d3888f6..f9daf04 100644
--- a/SUMMARY.md
+++ b/SUMMARY.md
@@ -5,6 +5,7 @@
   * [Heaps of Knowledge \[420 points\]](/binary-exploitation/heaps-of-knowledge-420-points.md)
 * [Cryptography](cryptography.md)
   * [RSA 1 \[50 points\]](/cryptography/rsa-1-50-points.md)
+  * [RSA 2 \[80 points\]](/cryptography/rsa-2-80-points.md)
   * [Hash on Hash \[100 points\]](/cryptography/hash-on-hash-100-points.md)
   * [Security Through Obscurity \[150 points\]](/cryptography/security-through-obscurity-150-points.md)
   * [RSA 4 \[200 points\]](/cryptography/rsa-4-200-points.md)
diff --git a/cryptography.md b/cryptography.md
index 01609af..fdb2507 100644
--- a/cryptography.md
+++ b/cryptography.md
@@ -6,7 +6,7 @@ This category focuses on using advanced mathematical topics to encrypt data to p
 * Flip My Letters \[20 points\]
 * [RSA 1 \[50 points\]](/cryptography/rsa-1-50-points.md)
 * Let Me Be Frank \[75 points\]
-* RSA 2 \[80 points\]
+* [RSA 2 \[80 points\]](/cryptography/rsa-2-80-points.md)
 * Decode Me \[100 points\]
 * [Hash on Hash \[100 points\]](/cryptography/hash-on-hash-100-points.md)
 * RSA 3 \[135 points\]
diff --git a/cryptography/rsa-2-80-points.md b/cryptography/rsa-2-80-points.md
new file mode 100644
index 0000000..6d4058a
--- /dev/null
+++ b/cryptography/rsa-2-80-points.md
@@ -0,0 +1,46 @@
+# RSA 2 - 80 points
+
+Some more RSA! This time, there's no P and Q... [this](https://github.com/EasyCTF/easyctf-2017-problems/blob/master/rsa2/ciphertext2.txt). 
+
+```
+n: 360863815763347129786223223753677186553479
+e: 65537
+c: 212985665370717183529999720499957413167571
+```
+
+### Solution
+
+RSA 2 is like [RSA 1](rsa-1-50-points.md), with the difference that this time we don't have the factorization of *N*.
+*N* is small enough, though, that it's easy to factor.
+For example, https://www.wolframalpha.com/input/?i=prime+factorization+of+360863815763347129786223223753677186553479
+gives us:
+
+ * *p* = 595630980471473199937
+ * *q* = 605851320019829172167
+
+Now we can just reuse any decryption code from [RSA 1](rsa-1-50-points.md)
+to find the private key *d*:
+
+ * *d* = *e*<sup>−1</sup> mod (*p*−1)(*q*−1) = 300201340241660215082768288814375000466433
+
+Using *d*, we decrypt the ciphertext as
+
+ * *M* = *c*<sup>*d*</sup> mod *N* = 136143999223212922673501593241914978429
+
+In hexadecimal,
+
+ * *M* = 0x666c61677b6c30775f6e5f616438307d
+
+Interpreting each pair of hexadecimal digits as an ASCII character
+(`66`→'f', `6c`→'l', etc.) reveals the flag:
+
+```
+flag{l0w_n_ad80} 
+```
+
+After the competition, the
+[grader.py script](https://github.com/EasyCTF/easyctf-2017-problems/blob/master/rsa2/grader.py)
+showed that flags were perhaps randomly generated
+as `l0w_n_` followed by a four-character random string.
+
+### External Writeups