What is the smallest number that, when divided by 35, 56 and 91 leaves remainders of 7 in each case?

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3240

3620

3647

3637

answer is B.

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Detailed Solution

Given numbers 35, 56 and 91.

2 35, 56, 91 2 35, 28, 91 2 35, 14, 91 5 35, 7, 91 7 7, 7, 13 13 1, 1, 13 1,1,1

LCM = 2 \times 2 \times 2 \times 5 \times 7 \times 13

\Rightarrow LCM = 3640

The number when divided should leave 7 as a reminder. Thus,
Required number

= LCM (35, 56, 91) + 7

\Rightarrow 3640 + 7

\Rightarrow 3647

Therefore, the smallest number that, when divided by 35, 56 and 91 leaves a remainder of 7 in each case is 3647.
Hence, option 2 is correct.

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