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Home » Question: How many positive 5-digit numbers are divisible by both $ 4 $ and $ 9 $? - Imagemakers

Question: How many positive 5-digit numbers are divisible by both $ 4 $ and $ 9 $? - Imagemakers

📅 May 2, 2026 👤 scraface

May 02, 2026

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Using the Pythagorean theorem, \( a^2 + b^2 = c^2 \), where \( c \) is...
Let \( a = 9 \), and \( c = 15 \). So, \( 9^2 + b^2 = 15^2 \).
That simplifies to \( 81 + b^2 = 225 \). Subtracting 81 from both side...
Taking the square root, \( b = 12 \) cm.
#### The other leg is 12 cm.
Solve for x: \( 3x^2 - 12x + 9 = 0 \).
The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
Here, \( a = 3 \), \( b = -12 \), \( c = 9 \).
Calculate the discriminant: \( b^2 - 4ac = (-12)^2 - 4 \times 3 \times...
So, \( x = \frac{12 \pm \sqrt{36}}{6} = \frac{12 \pm 6}{6} \).

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