Among the given numbers below, the smallest number which will be divided by 9, 10, 15 and 20, leaves the remainders 4, 5, 10, and 15, respectively

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: \( \boxed{355} \)