The smallest no. which one leaves remainder8 and 12 when divided by 28 and 32 respectively is
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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. = 224
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