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# A number lying between 1000 and 2000 is such that on division by 2, 3, 4, 5, 6, 7 and 8 leaves remainder respectively 1, 2, 3, 4, 5, 6 and 7. The number is

 A) 1876 B) 1679 C) 1778 D) 1654

 B) 1679

Note that =

=$$\Large \left(2-1\right)= \left(3-2\right)= \left(4-3\right)$$ $$\Large = \left(5-4\right)= \left(6-5\right)= \left(7-6\right)= \left(8-7\right)=1$$

LCM of 2, 3, 4, 5, 6, 7, 8 is 800

Required number = (840k-1)

Since the number lies between 1000 and 2000, the value of k=2

The number is 1680 - 12 = 1679.

Part of solved LCM and HCF questions and answers : >> Elementary Mathematics >> LCM and HCF

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