find the smallest number which when divided by 25,40,60, leaves remainder 7 in each case .

Hi!

Here is the answer to your question.

The given numbers are 25, 40 and 80.

The smallest number which when divided by 25, 40 and 60 leaves remainder 7 is obtained by adding 7 to the LCM of 25, 40 and 60.

Prime factorization of 25 = 5 × 5

Prime factorization of 40 = 2 × 2 × 2 × 5

Prime factorization of 60 = 2 × 2 × 3 × 5

LCM of 25, 40 and 60 = 2 × 2 × 2 × 3 × 5 × 5 = 600

∴Smallest number which when divided by 25, 40 and 60 leaves remainder 7 = 600 + 7 = 607

Cheers!

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