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Home » But wait: is it divisible by $9$ always? No, as $105 = 3 \cdot 5 \cdot 7$, not divisible by 9. - AMAZONAWS

But wait: is it divisible by $9$ always? No, as $105 = 3 \cdot 5 \cdot 7$, not divisible by 9. - AMAZONAWS

Does Every Number Divisible by 9 Always Work? No—Take 105, Which Proves the Point!

📅 March 11, 2026 👤 scraface
Mar 11, 2026
But wait: is it divisible by $9$ always? No, as $105 = 3 \cdot 5 \cdot 7$, not divisible by 9.

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📌 Solution: For a right triangle with legs $ a $ and $ b $, and hypotenuse $ z $, the inradius is given by $ c = rac{a + b - z}{2} $. The area $ A $ of the triangle is $ rac{1}{2}ab $. The area of the incircle is $ \pi c^2 $. Using the identity $ a^2 + b^2 = z^2 $, and expressing $ a + b $ in terms of $ z $ and $ c $, we find $ a + b = z + 2c $. Also, the semiperimeter $ s = rac{a + b + z}{2} = rac{2z + 2c}{2} = z + c $. The area can also be written as $ A = r \cdot s = c(z + c) $. Therefore, the ratio becomes:
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