2024 AMC 8 Problems/Problem 15: Difference between revisions

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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

@@ Line 9: / Line 9: @@
 ==Solution 1==
-The highest that <math>FLYFLY</math> can be would have to be <math>124124</math>, and it cannot be higher than that because then it would exceed the <math>6</math>-digit limit set on <math>BUGBUG</math>.
+The highest that <math>FLYFLY</math> can be would have to be <math>124124</math>, and it cannot be higher than that, because then it would be <math>125125</math>, and <math>125125</math> multiplied by 8 is <math>1000000</math>, and then it would exceed the <math>6</math> - digit limit set on <math>BUGBUG</math>.
-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 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>.
@@ Line 21: / Line 21: @@
 ~ cxsmi (minor formatting edits)
+~Alice of Evergreen Middle School
 ==Solution 2==

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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