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

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==Solution==
 
==Solution==
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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>.  
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The highest that FLYFLY can be would have to be 124124, and cannot exceed that because it would exceed the 6-digit limit set on BUGBUG.  
  
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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>.  
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So, if we start at 124124*8, we get 992992, which would be wrong because the numbers cannot be repeated.  
  
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Therefore, we have the following equation:
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If we move on to 123123 and multiply by 8, we get 984984, all the digits are different, so FLY+BUG would be 123+984, which is 1107. So, therefore, the answer is C, 1107.
  
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<math>8 \times 1001(\underline{F}~\underline{L}~\underline{Y}) = 1001(\underline{B}~\underline{U}~\underline{G})</math>.
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-Akhil Ravuri, John Adams Middle School
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Simplifying the equation gives
 
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<math>8(\underline{F}~\underline{L}~\underline{Y}) = (\underline{B}~\underline{U}~\underline{G})</math>.
 
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We can now use our equation to test each answer choice.
 
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We have that <math>123123 \times 8 = 984984</math>, so we can find the sum:
 
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<math>\underline{F}~\underline{L}~\underline{Y}+\underline{B}~\underline{U}~\underline{G} = 123 + 984= 1107</math>.
 
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So, the correct answer is <math>\textbf{(C)}\ 1107</math>.
 
  
 
==Video Solution by Math-X (First fully understand the problem!!!)==
 
==Video Solution by Math-X (First fully understand the problem!!!)==

Revision as of 13:48, 25 January 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

The highest that FLYFLY can be would have to be 124124, and cannot exceed that because it would exceed the 6-digit limit set on BUGBUG.

So, if we start at 124124*8, we get 992992, which would be wrong because the numbers cannot be repeated.

If we move on to 123123 and multiply by 8, we get 984984, all the digits are different, so FLY+BUG would be 123+984, which is 1107. So, therefore, the answer is C, 1107.

-Akhil Ravuri, John Adams Middle School

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

https://www.youtube.com/watch?v=JK4HWnqw-t0

~Math-X

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