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Q.

[[1]] is the smallest number that is divisible by 24, 36 and 54, and leaves a remainder of 5 every time.


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Detailed Solution

221 is the smallest number that is divisible by 24, 36 and 54, and leaves a remainder of 5 every time.
We have been given the numbers 24, 36 and 54.
The remainder = 5.
Calculating the prime factors of 24, 36 and 54,
24 = 2×2×2×3  
36 = 2×2×3×3  
54 = 2×3×3×3  
Calculating the L.C.M.,
⇒2×2×2×3×3×3
⇒216
So, 216 is the smallest number divisible by 24,36, and 54 leaving remainder 0.
Adding the remainder in the LCM to calculate the value of required number,
⇒216+5
⇒221
Hence, the required number is 221.
 
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