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Python/integer-break.py
@@ -37,7 +37,7 @@ def integerBreak(self, n):
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# 3^a * 2^b, s.t. 3a + 2b = n
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#
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# 2. To maximize the product = 3^a * 2^b s.t. 3a + 2b = n
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- # - For each b >= 3, we can always maximize the product by:
+ # - For each b >= 3, we can always maximize the product by:
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# 3^a * 2^b <= 3^(a+2) * 2^(b-3) s.t. 3(a+2) + 2(b-3) = n
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# Conclusion 2:
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