2024 AMC 8 Problems/Problem 15: Difference between revisions

Revision as of 15:01, 9 November 2024

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

Video Solution by WhyMath

https://youtu.be/GJcAdyDYpQk

@@ Line 84: / Line 84: @@
 ==Video Solution by Dr. David==
 https://youtu.be/kMkps16-xwE
+==Video Solution by WhyMath==
+https://youtu.be/GJcAdyDYpQk
 ==See Also==

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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