What would be the smallest natural number which when divided either by 20 or by 42 or by 76 leaves a remainder of 7 in each case?

3047

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6047

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7987

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63847

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Solution

The correct option is C
7987
Number is divided either by 20 or by 42 or by 76

= K × LCM (20, 42, 76) + constant difference

= 7980 K + 7 ( where K is natural number)

So, if we put K = 1 then,

Least number will be 7980 + 7 = 7987

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