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

(Solution 1)
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So, if we start at <math>124124\cdot8</math>, we get <math>992992</math>, which would be wrong because both <math>B \& U</math> would be <math>9</math>, and the numbers cannot be repeated between different letters.  
 
So, if we start at <math>124124\cdot8</math>, we get <math>992992</math>, which would be wrong because both <math>B \& U</math> would be <math>9</math>, and the numbers cannot be repeated between different letters.  
  
If we move on to the next lowest, <math>123123</math>, and multiply by <math>8</math>, we get <math>984984</math>. All the digits are different, so <math>FLY+BUG</math> would be <math>123+984</math>, which is <math>1107</math>. So, the answer is <math>\boxed{\textbf{(C)}1107}</math>.  
+
If we move on to the next highest, <math>123123</math>, and multiply by <math>8</math>, we get <math>984984</math>. All the digits are different, so <math>FLY+BUG</math> would be <math>123+984</math>, which is <math>1107</math>. So, the answer is <math>\boxed{\textbf{(C)}1107}</math>.  
  
 
- Akhil Ravuri of John Adams Middle School
 
- Akhil Ravuri of John Adams Middle School

Revision as of 08:51, 22 February 2024

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 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 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

Video Solution by Interstigation

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

See Also

2024 AMC 8 (ProblemsAnswer KeyResources)
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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