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Question
find the smallest number which when divided by 25, 40 , 60 leaves remainder 7 in each case .
find the smallest number which when divided by 25, 40 , 60 leaves remainder 7 in each case .
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Solution
The given numbers are 25, 40 and 60.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 × 5Prime factorization of 40 = 2 × 2 × 2 × 5Prime factorization of 60 = 2 × 2 × 3 × 5LCM 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
The given numbers are 25, 40 and 60.
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
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