This question is from the first chapter of class 10th NCERT chapter name is Real Number. Here in this question, we have to find the smallest number which when divided by 28 and 32 leaves the remainder 8 and 12 respectively. This is a very tricky question i was stuck in this question. Don’t know how to solve them, Please someone help me.

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# Find the smallest number which when divided by 28 and 32 leaves remainder 8 and 12 respectively.

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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.

Prime factorization of 28 = 2 x 2 x 7

Prime factorization of 32 = 2 x 2 x 2 x 2 x 2

LCM(28, 32) = 2 x 2 x 2 x 2 x 2 x 7 = 224.

Therefore, the required the smallest number = 224 – 20 = 204.