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Question
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
🎥 Video solution / Text Solution of this question is given below:
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: \( \boxed{355} \)
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