Difference between revisions of "2024 AMC 8 Problems/Problem 15"
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So, the correct answer is <math>\boxed{\textbf{(C)}\ 1107}</math>. | So, the correct answer is <math>\boxed{\textbf{(C)}\ 1107}</math>. | ||
| − | - C. | + | - C. Re |
==Solution 3 (Answer Choices)== | ==Solution 3 (Answer Choices)== | ||
| Line 53: | Line 53: | ||
~andliu766 | ~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!!!)== | ==Video Solution by Math-X (Apply this simple strategy that works every time!!!)== | ||
Latest revision as of 17:18, 8 December 2024
Contents
- 1 Problem 15
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3 (Answer Choices)
- 5 Solution 4
- 6 Video Solution by Math-X (Apply this simple strategy that works every time!!!)
- 7 Video Solution (A Clever Explanation You’ll Get Instantly)
- 8 Video Solution 2 (easy to digest) by Power Solve
- 9 Video Solution 3 (2 minute solve, fast) by MegaMath
- 10 Video Solution 4 by SpreadTheMathLove
- 11 Video Solution by NiuniuMaths (Easy to understand!)
- 12 Video Solution by CosineMethod
- 13 Video Solution by Interstigation
- 14 Video Solution by Dr. David
- 15 Video Solution by WhyMath
- 16 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 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. 
