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Which is the smallest number which leaves remainder 8 and 12 when divided by 28 and 32 respectively?

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Answer:

Which is the smallest number which leaves remainder 8 and 12 when divided by 28 and 32 respectively? 204.

Step-by-step explanation:

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Answer:

204 is the smallest number.

Step-by-step explanation:

28 - 8 = 20 and 32 - 12 = 20

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 x 2 x 2 x 2 x 2 x 7 = 224

Hence, the required smallest number,

→ 224 - 20 = 204

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