diff --git a/data_structures/hashing/number_theory/chinese_remainder_theorem.py b/data_structures/hashing/number_theory/chinese_remainder_theorem.py
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+++ b/data_structures/hashing/number_theory/chinese_remainder_theorem.py
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+# Chinese Remainder Theorem:
+
+# If GCD(a,b) = 1, then for any remainder ra modulo a and any remainder rb modulo b there exists integer n,
+# such that n = ra (mod a) and n = ra(mod b).  If n1 and n2 are two such integers, then n1=n2(mod ab)
+
+# Algorithm :
+
+# 1. Use extended euclid algorithm to find x,y such that a*x + b*y = 1
+# 2. Take n = ra*by + rb*ax
+
+
+# Extended Euclid
+def ExtendedEuclid(a, b):
+    if b == 0:
+        return (1, 0)
+    (x, y) = ExtendedEuclid(b, a % b)
+    k = a // b
+    return (y, x - k * y)
+
+
+# Uses ExtendedEuclid to find inverses
+def ChineseRemainderTheorem(n1, r1, n2, r2):
+    (x, y) = ExtendedEuclid(n1, n2)
+    m = n1 * n2
+    n = r2 * x * n1 + r1 * y * n2
+    return ((n % m + m) % m)
+
+
+# ----------SAME SOLUTION USING InvertModulo instead ExtendedEuclid----------------
+
+# This function find the inverses of a i.e., a^(-1)
+def InvertModulo(a, n):
+    (b, x) = ExtendedEuclid(a, n)
+    if b < 0:
+        b = (b % n + n) % n
+    return b
+
+
+# Same a above using InvertingModulo
+def ChineseRemainderTheorem2(n1, r1, n2, r2):
+    x, y = InvertModulo(n1, n2), InvertModulo(n2, n1)
+    m = n1 * n2
+    n = r2 * x * n1 + r1 * y * n2
+    return (n % m + m) % m