How many integers are there from 500 to 2000 that leave a remainder of 2 on division by 5 and a remainder of 1 on division by 7?

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Solution

The correct option is D
43

The first term of the sequence will be 512.The common difference will be 35.

512+(n-1)35 = 1982.

n-1 =42.

n = 43

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