Question

find the smallest number which leaves remainders 8 and 12 when divided by 18 and 32​

Answer 1

Answer:

The smallest number when divided by 28 and 32 leaves remainder 0 will be the LCM

 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.

224 = 28 × 8 +0

224 = 224 +0

224 = 216 + 8 ( because we want r = 8)

224 = (28*7+20)+8 ( since divisor given is 28)

224–20 = 28*7 +8 .............(1)

Similarly, 224 = 32 × 7 +0

224 = 224 +0

224 = 212 +12 ( because we want r = 12)

224 = (32×6+20) +12 ( since divisor given is 32)

224 - 20 = 32×6 +12 .............(2)

LHS of both eq (1) & eq(2) are equal. Divisors are also 28 & 32 & remainders are also the required numbers 8 & 12

So the smallest dividend = 224 - 20 = 204

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Answer 2

Answer:

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.

224 = 28 × 8 +0

224 = 224 +0

224 = 216 + 8 ( because we want r = 8)

224 = (28*7+20)+8 ( since divisor given is 28)

224–20 = 28*7 +8 .............(1)

Similarly, 224 = 32 × 7 +0

224 = 224 +0

224 = 212 +12 ( because we want r = 12)

224 = (32×6+20) +12 ( since divisor given is 32)

224 - 20 = 32×6 +12 .............(2)

LHS of both eq (1) & eq(2) are equal. Divisors are also 28 & 32 & remainders are also the required numbers 8 & 12

So the smallest dividend = 224 - 20 = 204