2024 AMC 8 Problems/Problem 15

1 Problem 15
2 Solution 1
3 Solution 2
4 Solution 3 (Answer Choices)
5 Video Solution (A Clever Explanation You’ll Get Instantly)
6 Video Solution 2 (easy to digest) by Power Solve
7 Video Solution 3 (2 minute solve, fast) by MegaMath
8 Video Solution 4 by SpreadTheMathLove
9 Video Solution by NiuniuMaths (Easy to understand!)
10 Video Solution by CosineMethod
11 Video Solution by Interstigation
12 Video Solution by Dr. David
13 See Also

Problem 15

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 be higher than that, because then it would be $125125$ , and $125125$ multiplied by 8 is $1000000$ , 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 $\boxed{\textbf{(C)}1107}$ .

- Akhil Ravuri of John Adams Middle School

- Aryan Varshney of John Adams Middle School (minor edits... props to Akhil for the main/full answer :D)

~ cxsmi (minor formatting edits)

~Alice of Evergreen Middle School

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 $\boxed{\textbf{(C)}\ 1107}$ .

- C. Ren

Solution 3 (Answer Choices)

Note that $FLY+BUG = 9 \cdot FLY$ . Thus, we can check the answer choices and find $FLY$ through each of the answer choices, we find the 1107 works, so the answer is $\boxed{\textbf{(C)}\ 1107}$ .

~andliu766

sus!

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/5ZIFnqymdDQ?si=rxqPhk-xiKjmbhNF&t=1683

~hsnacademy

Video Solution 2 (easy to digest) by Power Solve

https://youtu.be/TKBVYMv__Bg

Video Solution 3 (2 minute solve, fast) by MegaMath

https://www.youtube.com/watch?v=QvJ1b0TzCTc

Video Solution 4 by SpreadTheMathLove

https://www.youtube.com/watch?v=RRTxlduaDs8

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=V-xN8Njd_Lc

~NiuniuMaths

Video Solution by CosineMethod

https://www.youtube.com/watch?v=77UBBu1bKxk don't recommend but its quite clean, learn what you must- Orion 2010 minor edits by Fireball9746

Video Solution by Interstigation

https://youtu.be/ktzijuZtDas&t=1585

Video Solution by Dr. David

https://youtu.be/kMkps16-xwE

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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