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 exceed the <math>6</math>-digit limit set on <math>BUGBUG</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 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</math> & <math>U</math> would be <math>9</math>, and the numbers cannot be repeated. | + | So, if we start at <math>124124\cdot8</math>, we get <math>992992</math>, which would be wrong because both <math>B</math> & <math>U</math> would be <math>9</math>, and the numbers cannot be repeated between different letters. |
If we move on to the next lowest, <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>. | If we move on to the next lowest, <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>. | ||
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- Akhil Ravuri of John Adams Middle School | - Akhil Ravuri of John Adams Middle School | ||
− | ~ Aryan Varshney of John Adams Middle School (minor edits; props to Akhil Ravuri for the full solution :D) | + | ~ Aryan Varshney of John Adams Middle School (minor edits including grammar; props to Akhil Ravuri for the full solution :D) |
~ cxsmi (minor formatting edits) | ~ cxsmi (minor formatting edits) |
Revision as of 22:12, 26 January 2024
Contents
[hide]- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3(Answer choices)
- 5 Video Solution 1 (easy to digest) by Power Solve
- 6 Video Solution 2 by Math-X (First fully understand the problem!!!)
- 7 Video Solution 3 (2 minute solve, fast) by MegaMath
- 8 Video Solution 4 by SpreadTheMathLove
- 9 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 10 See Also
Problem
Let the letters ,,,,, represent distinct digits. Suppose is the greatest number that satisfies the equation
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 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 lowest, , 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 including grammar; props to Akhil Ravuri for the full solution :D)
~ cxsmi (minor formatting edits)
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. Ren, Thomas Grover Middle School
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
Video Solution 1 (easy to digest) by Power Solve
Video Solution 2 by Math-X (First fully understand the problem!!!)
https://www.youtube.com/watch?v=JK4HWnqw-t0
~Math-X
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 CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=77UBBu1bKxk
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.