Find the smallest number which when divided by 9, 10, 15 and 20 leaves remainders 4, 5, 10 and 15 respectively.

✅ Solution:

Let the number be \( x \).

• \( x \equiv 4 \mod 9 \Rightarrow x - 4 \) divisible by 9
• \( x \equiv 5 \mod 10 \Rightarrow x - 5 \) divisible by 10
• \( x \equiv 10 \mod 15 \Rightarrow x - 10 \) divisible by 15
• \( x \equiv 15 \mod 20 \Rightarrow x - 15 \) divisible by 20

So, \( x + 5 \) is divisible by LCM of 9, 10, 15, 20

LCM = \( 2^2 \cdot 3^2 \cdot 5 = 180 \)

\( x + 5 = 180 \times 2 = 360 \Rightarrow x = 355 \)

? Final Answer: \( \boxed{355} \)