2024 AMC 8

Complete problem set with solutions and individual problem pages

Problem 15 Medium

Let the letters F,L,Y,B,U,G represent distinct digits. Suppose \underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y} is the greatest number that satisfies the equation

8\cdot\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}=\underline{B}~\underline{U}~\underline{G}~\underline{B}~\underline{U}~\underline{G}.

What is the value of \underline{F}~\underline{L}~\underline{Y}+\underline{B}~\underline{U}~\underline{G}?

  • A.

    1089

  • B.

    1098

  • C.

    1107

  • D.

    1116

  • E.

    1125

Answer:C

Solution 1

The highest that FLYFLY can be would have to be 124124, and it cannot be higher than that, because then it would be 125125, and 125125 multiplied by 8 is 1001000, and then it would exceed the 6 - digit limit set on BUGBUG.

So, if we start at 124124\cdot8, we get 992992, which would be wrong because both B \& U would be 9, and the numbers cannot be repeated between different letters.

If we move on to the next highest, 123123, and multiply by 8, we get 984984. All the digits are different, so FLY+BUG would be 123+984, which is 1107. So, the answer is \boxed{\textbf{(C)}1107}.

 

Solution 2

Notice that \underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y} = 1000(\underline{F}~\underline{L}~\underline{Y}) + \underline{F}~\underline{L}~\underline{Y}.

Likewise, \underline{B}~\underline{U}~\underline{G}~\underline{B}~\underline{U}~\underline{G} = 1000(\underline{B}~\underline{U}~\underline{G}) + \underline{B}~\underline{U}~\underline{G}.

Therefore, we have the following equation:

8 \times 1001(\underline{F}~\underline{L}~\underline{Y}) = 1001(\underline{B}~\underline{U}~\underline{G}).

Simplifying the equation gives

8(\underline{F}~\underline{L}~\underline{Y}) = (\underline{B}~\underline{U}~\underline{G}).

We can now use our equation to test each answer choice.

We have that 123123 \times 8 = 984984, so we can find the sum:

\underline{F}~\underline{L}~\underline{Y}+\underline{B}~\underline{U}~\underline{G} = 123 + 984 = 1107.

So, the correct answer is \boxed{\textbf{(C)}\ 1107}.