Difference between revisions of "2024 AMC 8 Problems/Problem 15"
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~ 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; props to Akhil Ravuri for the full solution :D) | ||
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~ cxsmi (minor formatting edits) | ~ cxsmi (minor formatting edits) | ||
Revision as of 16:26, 26 January 2024
Contents
[hide]- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Video Solution 1 by Math-X (First fully understand the problem!!!)
- 5 Video Solution 2 (easy to digest) by Power Solve
- 6 Video Solution 3 (2 minute solve, fast) by MegaMath
- 7 Video Solution 4 by SpreadTheMathLove
- 8 Video Solution by CosineMethod [🔥Fast and Easy🔥]
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 exceed that because 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.
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; 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
Video Solution 1 by Math-X (First fully understand the problem!!!)
https://www.youtube.com/watch?v=JK4HWnqw-t0
~Math-X
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