2023 AMC 12B Problem 16

Problem 16 of 25IntermediateNumber Theory

In Coinland, there are three types of coins, each worth

6,

10,

and

15.

What is the sum of the digits of the maximum amount of money that is impossible to have?

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Solution

The amounts

30,31,32,33,34,35

are all attainable (for instance

30=6\cdot 5,

31=6+10+15,

32=6\cdot 2+10\cdot 2,

33=6\cdot 3+15,

34=6\cdot 4+10,

35=10+10+15

). Adding

6

’s then reaches every larger amount. Checking below,

29

is impossible: without a

15

-coin a sum of

6

’s and

10

’s is even, while with one

15

-coin the remaining

14

cannot be written as

6a+10b;

two

15

-coins already exceed

29.

So the largest impossible amount is

29,

whose digit sum is

2+9=11.

Thus, the correct answer is D.