A man wishes to sell a puppy for $11. A customer wants to buy it but only has foreign currency. The exchange rate for the foreign currency is 11 round coins = $15, 11 square coins = $16, 11 triangular coins = $17.
How many of each coinage should the customer pay?
By the problem, the respective cost of each round coin, each square coin and each triangular coin are $15/11, $16/11 and $17/11.
Let the respective number of round coin(s), square coin(s) and triangular coin(s) be r, s and t.
Then, we must have:
(15r+16s+17t)/11 =11
=> 15r+16s+17t =121
Now, dividing both sides by 16, we have:
r+s+t+(t-r)/16= 121/16
Now, there are ONLY two cases involving proper fractions, and these are :
r+s+t +(t-r)/16= (7+9/16) OR (8-7/16)
CASE 1: r+s+t+(t-r)/16= 7+9/16
=> r+s+t=7 and, t-r=9
Subtracting, we have: 2r+s=-2. The rhs of this equation is negative, and this is a contradiction as r and s are non negative integers
CASE 2: r+s+t+(t-r)/16 = 8 - 7/16
=> r+s+t= 8 and, r-t=7
Subtracting, we have: s+2t=1, whose only nonnegative integer solution occurs at:
(s,t)=(1,0), giving: r =8-1-0=7
Consequently, the customer should pay precisely 7 Round Coins and 1 Square Coin.
Edited on September 30, 2023, 8:21 am
Puzzle Solution
