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What is the smallest number that is divisible by 35, 56, and 91, and leaves remainders of 7 every time?

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3650

3670

3647

3435

answer is C.

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Detailed Solution

We have been the numbers 35, 56, and 91.
The prime factorization of 35, 56 and 91,

35 = 5 \times 7

56 = 2 × 2 × 2 × 7

91 = 7 × 13

LCM of 35, 56 and 91 is 3640.
Adding the remainder in the LCM,

\Rightarrow 3640 + 7 = 3647

Therefore, the required number is 3647.
Hence, option 3 is correct.

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