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 | + | 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. |
Revision as of 17:32, 26 January 2024
Contents
- 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 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.
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