Find the greatest 4digit number which when divided by 20,24and 45 leaves a remainder of 18 in each case

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   prajapatyk01

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   Given numbers,

   20 , 24 and 45

   Prime factorization of

   20=2²×5

   24=2³×3

   45=3²×5

   LCM=product of each prime factor of highest power

   LCM=2³×3²×5=360

   Greatest four digit number=9999

   greatest four digit number divisible by all given numbers=9999-remainder when 9999 is divided by LCM of given numbers

   greatest four digit number divisible by given numbers=9999-279=9720

   Given that,

   required number when divided by 20 , 24, 45 leaves remainder 18.

   Therefore,required number=9720+18=9738

   Hence greatest four digit number when divided by 20,24 and 45 is 9738.

   Step-by-step explanation:

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