diff --git a/extra/project-euler/087/087.factor b/extra/project-euler/087/087.factor
index c20498415c..0367e7a5a3 100644
--- a/extra/project-euler/087/087.factor
+++ b/extra/project-euler/087/087.factor
@@ -1,29 +1,53 @@
-USING: locals math math.primes sequences math.functions sets kernel ;
+USING: locals math math.functions math.primes
+project-euler.common sequences sets ;
+
 IN: project-euler.087
 
-<PRIVATE
+! https://projecteuler.net/index.php?section=problems&id=87
 
-: remove-duplicates ( seq -- seq' )
-    dup intersect ;
+! DESCRIPTION
+! -----------
+
+! The smallest number expressible as the sum of a prime square,
+! prime cube, and prime fourth power is 28. In fact, there are
+! exactly four numbers below fifty that can be expressed in such
+! a way:
+
+! 28 = 2^2 + 2^3 + 2^4
+! 33 = 3^2 + 2^3 + 2^4
+! 49 = 5^2 + 2^3 + 2^4
+! 47 = 2^2 + 3^3 + 2^4
+
+! How many numbers below fifty million can be expressed as the
+! sum of a prime square, prime cube, and prime fourth power?
+
+<PRIVATE
 
 :: prime-powers-less-than ( primes pow n -- prime-powers )
     primes [ pow ^ ] map [ n <= ] filter ;
 
-! You may think to make a set of all possible sums of a prime square and cube 
-! and then subtract prime fourths from numbers ranging from 1 to 'n' to find 
-! this. As n grows large, this is actually more inefficient!
+! You may think to make a set of all possible sums of a prime
+! square and cube and then subtract prime fourths from numbers
+! ranging from 1 to 'n' to find this. As n grows large, this is
+! actually more inefficient!
+!
 ! Prime numbers grow ~ n / log n
+!
 ! Thus there are (n / log n)^(1/2) prime squares <= n,
 !                (n / log n)^(1/3) prime cubes   <= n,
 !            and (n / log n)^(1/4) prime fourths <= n.
-! If we compute the cartesian product of these, this takes 
+!
+! If we compute the cartesian product of these, this takes
 ! O((n / log n)^(13/12)).
-! If we instead precompute sums of squares and cubes, and iterate up to n,
-! checking each fourth power against it, this takes
+!
+! If we instead precompute sums of squares and cubes, and
+! iterate up to n, checking each fourth power against it, this
+! takes:
+!
 ! O(n * (n / log n)^(1/4)) = O(n^(5/4)/(log n)^(1/4)) >> O((n / log n)^(13/12))
 !
-! When n = 50000000, the first equation is approximately 10 million and 
-! the second is approximately 2 billion.
+! When n = 50,000,000, the first equation is approximately 10
+! million and the second is approximately 2 billion.
 
 :: prime-triples ( n -- answer )
     n sqrt primes-upto                :> primes
@@ -32,9 +56,11 @@ IN: project-euler.087
     primes 4 n prime-powers-less-than :> primes^4
     primes^2 primes^3 [ + ] cartesian-map concat
              primes^4 [ + ] cartesian-map concat
-    [ n <= ] filter remove-duplicates length ;
+    [ n <= ] filter members length ;
 
 PRIVATE>
 
 :: euler087 ( -- answer )
-    50000000 prime-triples ;
+    50,000,000 prime-triples ;
+
+SOLUTION: euler087