Question

find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively . please answer with explanation clearly .....​

Answer 1

when 28 is divided by the number then it will remainder 8 . so the completely divisible number is 28-8=20

and when 32 divided by the number then it will remainder 12 so the completely divisible number os 32-12=20 .

there the number which will leave remainder 8 and 12 when divided by 28 and 32 is 20.

Answer 2

let the number be x

x = lcm of 28 and 32 - hcf of 28-8 and 32-12

x = (lcm of 28 and 32) - 20

x = 224-20

(hcf and lcm explanation is in the attachment )

So x = 204

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