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Q.

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

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a

3240

b

3620 

c

3647

d

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=2×2×2×5×7×13
⇒LCM=3640
The number when divided should leave 7 as a reminder. Thus,
Required number =LCM(35,56,91)+7
⇒ 3640+7
⇒ 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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What is the smallest number that, when divided by 35, 56 and 91 leaves remainders of 7 in each case?