2024 AMC 8 Problems/Problem 15

Problem

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 $\underline{F}~\underline{L}~\underline{Y}+\underline{B}~\underline{U}~\underline{G}$ ?

$\textbf{(A)}\ 1089 \qquad \textbf{(B)}\ 1098 \qquad \textbf{(C)}\ 1107 \qquad \textbf{(D)}\ 1116 \qquad \textbf{(E)}\ 1125$

Solution 1

The highest that $FLYFLY$ can be would have to be $124124$ , and it cannot exceed that because 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 the numbers cannot be repeated.

If we move on to $123123$ 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 $\boxed{\textbf{(C)}1107}$ .

-Akhil Ravuri, John Adams Middle School ~ cxsmi (minor formatting edits)

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