One of the tribesmen had 3 gold, 5 silver and 8 bronze coins. He made some exchanges and ended up with least number of total coins. How many coins does he have now?
Explanation:
Consider the solution to first question of this set. 1 gold coin = 10 bronze coins and 1 silver coin = 4 bronze coins.
The tribesman initially had 3 gold + 5 silver + 8 bronze coins
Value of these coins in terms of bronze coins = 3 × 10b + 5 × 4b + 8b = 58 bronze coins.
To minimize the total number of coins such that their value (in terms of bronze coins) is 58 he must have maximum possible number of gold coins and then maximum possible number of silver coins and then remaining coins can be bronze coins.
At the end maximum gold coins, he can have is 5 and then maximum silver coins he can have is 2.
∴ The least number of coins he can have = 5 + 2 = 7 coins.
Hence, 7.
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