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Question

Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.


Solution

Given that the smallest number when divided by 28 and 32 leaves remainder 8 and 12 respectively.

28 - 8 = 20 and 32 - 12 = 20 are divisible by the required numbers.


Therefore the required number will be 20 less than the LCM of 28 and 32.

Prime factorization of 28 = 2 * 2 * 7

Prime factorization of 32 = 2 * 2 * 2 * 2 * 2

LCM(28,32) = 2 * 2 * 2 * 2 * 2 * 7

                     = 224.


Therefore the required smallest number = 224 - 20

                                                                    = 204.


Verification:

204/28 = 28 * 7 = 196.

             = 204 - 196   
 
             = 8


204/32 = 32 * 6 = 192

             = 204 - 192

             = 12.


Hope this helps!

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