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: $\boxed{355}$