GATE CS 2018 QGA-4 — General Aptitude

Let the required smallest natural number be $N$.

According to the problem, when $N$ is divided by $20$, $42$, or $76$, it leaves a remainder of $7$ in each case. This can be expressed as:

For the smallest such natural number, we set $k = 1$:

To find the $\text{LCM}(20, 42, 76)$, we find their prime factorizations:

• $20 = 2^2 \times 5$
• $42 = 2 \times 3 \times 7$
• $76 = 2^2 \times 19$

Taking the highest power of each prime factor:

Therefore, the required number is:

Hence, the correct option is C.