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

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==Problem==
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==Problem 15==
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Let the letters <math>F</math>,<math>L</math>,<math>Y</math>,<math>B</math>,<math>U</math>,<math>G</math> represent distinct digits. Suppose <math>\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}</math> is the greatest number that satisfies the equation
  
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==Solution==
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<cmath>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}.</cmath>
 +
 
 +
What is the value of <math>\underline{F}~\underline{L}~\underline{Y}+\underline{B}~\underline{U}~\underline{G}</math>?
 +
 
 +
<math>\textbf{(A)}\ 1089 \qquad \textbf{(B)}\ 1098 \qquad \textbf{(C)}\ 1107 \qquad \textbf{(D)}\ 1116 \qquad \textbf{(E)}\ 1125</math>
 +
 
 +
==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 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.
 +
 
 +
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
 +
 
 +
 
 +
- 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 <math>\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}</math>.
 +
 
 +
Likewise, <math>\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}</math>.
 +
 
 +
Therefore, we have the following equation:
 +
 
 +
<math>8 \times 1001(\underline{F}~\underline{L}~\underline{Y}) = 1001(\underline{B}~\underline{U}~\underline{G})</math>.
 +
 
 +
Simplifying the equation gives
 +
 
 +
<math>8(\underline{F}~\underline{L}~\underline{Y}) = (\underline{B}~\underline{U}~\underline{G})</math>.
 +
 
 +
We can now use our equation to test each answer choice.
 +
 
 +
We have that <math>123123 \times 8 = 984984</math>, so we can find the sum:
 +
 
 +
<math>\underline{F}~\underline{L}~\underline{Y}+\underline{B}~\underline{U}~\underline{G} = 123 + 984 = 1107</math>.
 +
 
 +
So, the correct answer is <math>\boxed{\textbf{(C)}\ 1107}</math>.
 +
 
 +
- C. Re
 +
 
 +
==Solution 3 (Answer Choices)==
 +
Note that <math>FLY+BUG = 9 \cdot FLY</math>. Thus, we can check the answer choices and find <math>FLY</math> through each of the answer choices, we find the 1107 works, so the answer is <math>\boxed{\textbf{(C)}\ 1107}</math>.
 +
 
 +
~andliu766
 +
 
 +
==Solution 4==
 +
To solve \( 8 \cdot FLY = BUG \) without guessing, we start by noting that \( FLY \) and \( BUG \) are three-digit numbers with distinct digits. We can write \( FLY = 100F + 10L + Y \) and \( BUG = 100B + 10U + G \), so the equation becomes:
 +
\[
 +
8 \cdot (100F + 10L + Y) = 100B + 10U + G.
 +
\]
 +
Since \( 8 \cdot FLY \) must also be a three-digit number, \( FLY \) must be small, specifically \( 100 \leq FLY \leq 124 \), because \( 8 \cdot 124 = 992 \) (just under 1000). We try \( F = 1 \) (since \( FLY \) must remain three digits), which gives \( B = 8F = 8 \), leading to \( FLY = 123 \) and \( BUG = 8 \cdot 123 = 984 \), where all digits \( 1, 2, 3, 9, 8, 4 \) are distinct. Adding \( FLY + BUG = 123 + 984 = 1107 \), the final answer is \( \boxed{1107} \).
 +
 
 +
\texttt{Sharpwhiz17}
 +
 
 +
==Video Solution by Math-X (Apply this simple strategy that works every time!!!)==
 +
https://youtu.be/BaE00H2SHQM?si=8CKIFksKvhOWABN3&t=3485
 +
 
 +
~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==
 +
 
 +
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
 +
 
 +
==See Also==
 +
{{AMC8 box|year=2024|num-b=14|num-a=16}}
 +
{{MAA Notice}}
 +
[[Category:Introductory Number Theory Problems]]

Latest revision as of 17:18, 8 December 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. Re

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

Solution 4

To solve \( 8 \cdot FLY = BUG \) without guessing, we start by noting that \( FLY \) and \( BUG \) are three-digit numbers with distinct digits. We can write \( FLY = 100F + 10L + Y \) and \( BUG = 100B + 10U + G \), so the equation becomes: \[ 8 \cdot (100F + 10L + Y) = 100B + 10U + G. \] Since \( 8 \cdot FLY \) must also be a three-digit number, \( FLY \) must be small, specifically \( 100 \leq FLY \leq 124 \), because \( 8 \cdot 124 = 992 \) (just under 1000). We try \( F = 1 \) (since \( FLY \) must remain three digits), which gives \( B = 8F = 8 \), leading to \( FLY = 123 \) and \( BUG = 8 \cdot 123 = 984 \), where all digits \( 1, 2, 3, 9, 8, 4 \) are distinct. Adding \( FLY + BUG = 123 + 984 = 1107 \), the final answer is \( \boxed{1107} \).

\texttt{Sharpwhiz17}

Video Solution by Math-X (Apply this simple strategy that works every time!!!)

https://youtu.be/BaE00H2SHQM?si=8CKIFksKvhOWABN3&t=3485

~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

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

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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