From ee9dc87ab4ca1a61a395e74a42758058d95ff7c0 Mon Sep 17 00:00:00 2001
From: Sidney Zhang <zly@lyzhang.me>
Date: Mon, 15 Dec 2025 17:29:36 +0800
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---
 solutions/0009/readme.md             |  7 ++++++
 solutions/0010.SumPrimes/eular_10.py | 34 ++++++++++++++++++++++++++++
 2 files changed, 41 insertions(+)
 create mode 100644 solutions/0009/readme.md
 create mode 100644 solutions/0010.SumPrimes/eular_10.py

diff --git a/solutions/0009/readme.md b/solutions/0009/readme.md
new file mode 100644
index 0000000..6edfb56
--- /dev/null
+++ b/solutions/0009/readme.md
@@ -0,0 +1,7 @@
+这个问题从数学上，需要换元替代计算。
+
+a = k(m^2 - n^2)
+b = 2kmn
+c = k(m^2 + n^2)
+
+基于这个换元进新推导。或者就是直接基于原始命题进行计算，可以参见 [这个解答](https://math.stackexchange.com/a/3378895) 。
diff --git a/solutions/0010.SumPrimes/eular_10.py b/solutions/0010.SumPrimes/eular_10.py
new file mode 100644
index 0000000..49a33d0
--- /dev/null
+++ b/solutions/0010.SumPrimes/eular_10.py
@@ -0,0 +1,34 @@
+"""
+Sum of primes below 2 million
+"""
+
+import time
+
+
+def timer(func):
+    def wrapper(*args, **kwargs):
+        start = time.time()
+        result = func(*args, **kwargs)
+        end = time.time()
+        print(f"{func.__name__} took {end - start:.6f} seconds")
+        return result
+
+    return wrapper
+
+
+def sieve_primes(max: int) -> list[int]:
+    primes = [True] * (max + 1)
+    primes[0:2] = [False, False]
+    for i in range(2, max + 1):
+        if primes[i]:
+            primes[i * i : max : i] = [False] * ((max - 1 - i * i) // i + 1)
+    return [i for i in range(max) if primes[i]]
+
+
+@timer
+def sum_primes(max: int) -> int:
+    return sum(sieve_primes(max))
+
+
+if __name__ == "__main__":
+    print(sum_primes(2_000_000))