Question

find find the smallest number which when divided by 28 and 32 leaves remainders 8 and 12 respectively ​

Answer 1

Solution:-

Given,

The smallest number when divided by 28 and 2 leaves remainder 8 and 12 respectively.

28 - 8  = 20

32 - 12 = 20

These numbers are divisible by the required Numbers.

Hence,

The smallest number be less than 20.

Now,

We should find the L.C.M. of 28 and 2

Prime Factorization of 28 = 2  * 2 * 7

Prime Factorization of 32 = 2 * 2 * 2 * 2 * 2

L.C.M. of 28 and 32 = 2 * 2 * 2 * 2 * 2 * 7

= 224

∴ The smallest number = 224 - 20

           

= 204.

Verification :

204/28 = 28 x 7 = 196.

            = 204 - 196    

              = 8

204/32 = 32 x 6 = 192

            = 204 - 192

            = 12.

∴ The smallest Number = 204.

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