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

(→‎Video Solution by Math-X (First fully understand the problem!!!))
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==Solution 1==
 
==Solution 1==
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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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The highest that <math>FLYFLY</math> can be would have to be <math>124124</math>, and it cannot exceed that because it would exceed the <math>6</math>-digit limit set on <math>BUGBUG</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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So, if we start at <math>124124\cdot8</math>, we get <math>992992</math>, which would be wrong because the numbers cannot be repeated.  
  
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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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If we move on to <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)}1007}</math>.  
  
 
-Akhil Ravuri, John Adams Middle School
 
-Akhil Ravuri, John Adams Middle School
 +
~ cxsmi (minor formatting edits)
  
 
==Solution 2==
 
==Solution 2==

Revision as of 14:28, 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 1

The highest that $FLYFLY$ can be would have to be $124124$, and it cannot exceed that because 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 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, the answer is $\boxed{\textbf{(C)}1007}$.

-Akhil Ravuri, John Adams Middle School ~ 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 $\textbf{(C)}\ 1107$.

- C. Ren, Thomas Grover Middle School

Video Solution by Math-X

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

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