Difference between revisions of "2024 AMC 8 Problems/Problem 15"
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| − | 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> | + | 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>1001000</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. | ||
Revision as of 22:04, 4 January 2025
Contents
- 1 Problem 15
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3 (Answer Choices)
- 5 Solution 4
- 6 Video Solution by Central Valley Math Circle (Goes through full thought process)
- 7 Video Solution by Math-X (Apply this simple strategy that works every time!!!)
- 8 Video Solution (A Clever Explanation You’ll Get Instantly)
- 9 Video Solution 2 (easy to digest) by Power Solve
- 10 Video Solution 3 (2 minute solve, fast) by MegaMath
- 11 Video Solution 4 by SpreadTheMathLove
- 12 Video Solution by NiuniuMaths (Easy to understand!)
- 13 Video Solution by CosineMethod
- 14 Video Solution by Interstigation
- 15 Video Solution by Dr. David
- 16 Video Solution by WhyMath
- 17 See Also
Problem 15
Let the letters 
,
,
,
,
,
represent distinct digits. Suppose 
is the greatest number that satisfies the equation
![\[8\cdot\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}=\underline{B}~\underline{U}~\underline{G}~\underline{B}~\underline{U}~\underline{G}.\]](http://latex.artofproblemsolving.com/0/4/7/047513b257239e4538e713105a012a223b79ded7.png)
What is the value of 
?
Solution 1
The highest that 
can be would have to be 
, and it cannot be higher than that, because then it would be 
, and multiplied by 8 is 
, and then it would exceed the 
- digit limit set on 
.
So, if we start at 
, we get 
, which would be wrong because both 
would be 
, and the numbers cannot be repeated between different letters.
If we move on to the next highest, 
, and multiply by 
, we get 
. All the digits are different, so 
would be 
, which is 
. So, the answer is 
.
- 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 
.
Likewise, 
.
Therefore, we have the following equation:
.
Simplifying the equation gives
.
We can now use our equation to test each answer choice.
We have that 
, so we can find the sum:
.
So, the correct answer is 
.
- C. Re
Solution 3 (Answer Choices)
Note that 
. Thus, we can check the answer choices and find 
through each of the answer choices, we find the 1107 works, so the answer is .
~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 Central Valley Math Circle (Goes through full thought process)
~mr_mathman
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
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
Video Solution by WhyMath
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. 
