Discussion
Three gold coins of weight 780gm, 840gm and 960gm are cut into small pieces, all of which have the equal weight. Each piece must be heavy as possible. If one such piece is shared by two persons, then how many persons are needed to give all the pieces of gold coins?
Answer: A
Every piece must weigh the same and be as heavy as possible, which is exactly the highest common factor of the three coins:
\[ 780 = 2^{2}\cdot3\cdot5\cdot13, \quad 840 = 2^{3}\cdot3\cdot5\cdot7, \quad 960 = 2^{6}\cdot3\cdot5 \]
\[ \operatorname{HCF} = 2^{2}\cdot3\cdot5 = 60 \text{ g} \]
Counting the pieces each coin yields:
\[ \frac{780}{60} + \frac{840}{60} + \frac{960}{60} = 13 + 14 + 16 = 43 \]
Each piece is shared by two people, so
\[ 43 \times 2 = 86 \text{ persons} \]
43 is the trap for stopping at the piece count and forgetting the sharing.
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