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Q.

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


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a

3650

b

3670

c

3647

d

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=5×7 56=2×2×2×7  
91=7×13    
LCM of 35, 56 and 91 is 3640.
Adding the remainder in the LCM,
⇒3640+7=3647
Therefore, the required number is 3647.
Hence, option 3 is correct.
 
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