Find the smallest number which when divided by 24,36 and 54 gives
remainder of 5 each time.
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Answer:
221.2
Step-by-step explanation:
Prime factorization of
24 = 2 × 2 × 2 × 3
36 = 2 × 2 × 3 × 3
54 = 2 × 3 × 3 × 3
So the required LCM = 2 × 2 × 2 × 3 × 3 × 3 = 216
The smallest number which is exactly divisible by 24, 36 and 54 is 216
In order to get remainder as 5
Required smallest number = 216 + 5 = 221
Therefore, the smallest number which when divided by 24, 26 and 54 gives a remainder of 5 each time is 221.
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