GATE | GATE CS 2018 | Question 2

What would be the smallest natural number which when divided either by 20 or by 42 or by 76 leaves a remainder ‘7’ in each case is_
(A) 3047
(B) 6047
(C) 7987
(D) 63847


Answer: (C)

Explanation: We need a number that can be written as 20x + 7, 42y + 7, 76z + 7 for three integers x, y and z.

We basically need to find least common multiple of 20 (2 * 2 *5), 42(2 * 3 * 7) and 76(2 * 2 * 19) which is

2 * 2 * 5 * 3 * 7 * 19 = 20 * 21 * 19 = 420 * 19 = 7980

So our number is 7980 + 7 = 7987.

Note – You can also verify options using virtual calculator.


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