一條數學題!急 20分

The answer is 23.

We can devided into 5 cases.

Let the amount of money paid be X,

Case 1: When the amount is divisible by 5.
Definitely we can paid by the coins.

Case 2: When the amount is divided by 5, the remainder is 1.
We can paid by 7*3 + 5*n, where n = (X-21)/5,
X >= 21 in this case.
Therefore, any amount having a remainder of 1 when divided by 5 and its amount is smaller than 21 cannot be paid by coins.
The maximum amount of money that cannot be paid by these combination is 16.

Similarly, Consider
Case 3: When the amount of remainder is 2 when divided by 5.
We can paid by 7 + 5*n where n = (X-7) DIV 5
X >= 7
The maximum amount of money that cannot be paid by these combination is 2.

Case 4: When the amount of remainder is 3 when divided by 5.
We can paid by 7*4 + 5*n where n = (X-28) DIV 5
X>=28
The maximum amount of money that cannot be paid by these combination is 23.

Case 5: When the amount of remainder is 4 when divided by 5.
We can paid by 7*2 + 5*n where n = (X-14) DIV 5
X >= 14
The maximum amount of money that cannot be paid by these combination is 9.

Combining the above results, we can find out the ansewr is 23.

ALTERNATIVE METHOD
You can also divided into 7 case.
Case 1: The remainder is 0 when divided by 7.
Case 2: The remainder is 1 when divided by 7.
Case 3: The remainder is 2 when divided by 7.
Case 4: The remainder is 3 when divided by 7.
Case 5: The remainder is 4 when divided by 7.
Case 6: The remainder is 5 when divided by 7.
Case 7: The remainder is 6 when divided by 7.

By similar method, we can also find the answer also.