Discussion
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?
Answer: C
The number is 7987.
A number leaving remainder 7 on division by 20, 42 and 76 is 7 more than a common multiple of all three. The smallest such number is 7 more than their least common multiple.
\[ \begin{aligned} 20 &= 2^{2} \cdot 5,\quad 42 = 2 \cdot 3 \cdot 7,\quad 76 = 2^{2} \cdot 19 \\ \text{LCM} &= 2^{2} \cdot 3 \cdot 5 \cdot 7 \cdot 19 = 7980 \end{aligned} \]
So the number is 7980 + 7 = 7987. 3047 and 6047 do not work: 3040 and 6040 are not multiples of 42 (or of 76). 63847 does leave remainder 7, since 63840 = 8 × 7980, but it is not the smallest.
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