find the smallest 3 digit number which when increased by 5 is divisible by 48 32 and 24

The given numbers are 48, 32 and 24.

At first, we will find the LCM of 48, 32 and 24.  

The prime factorisation of 48, 32 and 24 is :

48 = 2×2×2×2×332 = 2×2×2×2×224 = 2×2×2×3LCM of 48, 32 and 24 = 2×2×2×2×2×3 = 96The smallest 3-digit number divisible by 48, 32 and 24 = 96×2 = 192Required number = 192 - 5 = 187

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