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Difference between revisions of "2024 AMC 8 Problems/Problem 15"

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https://youtu.be/BaE00H2SHQM?si=4za1LGPg_w2gwosm&t=3484
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https://youtu.be/BaE00H2SHQM?si=4za1LGPg_w2gwosm&t=3484
 
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~Math-X
 
  
 
==Video Solution 2 (easy to digest) by Power Solve==
 
==Video Solution 2 (easy to digest) by Power Solve==

Revision as of 08:17, 6 September 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 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)

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

Video Solution 1 by Math-X (First fully understand the problem!!!)

https://youtu.be/BaE00H2SHQM?si=4za1LGPg_w2gwosm&t=3484

~Math-X

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 [🔥Fast and Easy🔥]

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

Video Solution by Interstigation

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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