Final answer:
To find the greatest possible number of coins Anna could have, set up an equation where Anna's coins plus 23 is equal to Willy's four times coins minus 23. Solve the equation to get that Anna could have a maximum of 15 coins before Willy gave her any.
Step-by-step explanation:
The problem states that Willy has at least four times as many coins as Anna has. If Willy gives Anna 23 coins, they will both have the same amount. Let's denote the number of coins Anna has as A. Therefore, Willy has at least 4A coins. After giving 23 coins to Anna, Willy has 4A - 23 coins, and Anna has A + 23 coins. Since they end up with the same amount, we can set these two expressions equal to each other to find the greatest possible number of coins Anna could have originally.
So we have A + 23 = 4A - 23
Moving variable terms to one side and constants to the other gives us:
3A = 46
Dividing both sides by 3 gives us:
A = 46 / 3
A = 15.333...
Since Anna cannot have a fraction of a coin, we round down to the nearest whole number:
A = 15
Thus, the greatest possible number of coins Anna could have is 15, which corresponds to option c.