Difference between revisions of "2024 AMC 8 Problems/Problem 15"
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− | ==Problem== | + | ==Problem 15== |
Let the letters <math>F</math>,<math>L</math>,<math>Y</math>,<math>B</math>,<math>U</math>,<math>G</math> represent distinct digits. Suppose <math>\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}</math> is the greatest number that satisfies the equation | Let the letters <math>F</math>,<math>L</math>,<math>Y</math>,<math>B</math>,<math>U</math>,<math>G</math> represent distinct digits. Suppose <math>\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}</math> is the greatest number that satisfies the equation | ||
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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 | + | 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. |
− | If we move on to the next | + | If we move on to the next highest, <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>. |
- 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 for the main/full answer :D) | ||
~ cxsmi (minor formatting edits) | ~ cxsmi (minor formatting edits) | ||
Line 46: | Line 47: | ||
==Solution 3 (Answer Choices)== | ==Solution 3 (Answer Choices)== | ||
− | Note that <math>FLY+BUG = 9 \cdot FLY</math>. Thus, we can check the answer choices and find <math>FLY</math> through each of the answer choices, we find the 1107 works, so the answer is <math>\textbf{(C)}\ 1107</math>. | + | Note that <math>FLY+BUG = 9 \cdot FLY</math>. Thus, we can check the answer choices and find <math>FLY</math> through each of the answer choices, we find the 1107 works, so the answer is <math>\boxed{\textbf{(C)}\ 1107}</math>. |
~andliu766 | ~andliu766 | ||
− | ==Video Solution 1 | + | ==Video Solution 1 by Math-X (First fully understand the problem!!!)== |
− | + | https://youtu.be/BaE00H2SHQM?si=4za1LGPg_w2gwosm&t=3484 | |
− | |||
− | |||
− | https:// | ||
~Math-X | ~Math-X | ||
− | == Video Solution by | + | ==Video Solution 2 (easy to digest) by Power Solve== |
− | + | https://youtu.be/TKBVYMv__Bg | |
− | https:// | ||
==Video Solution 3 (2 minute solve, fast) by MegaMath== | ==Video Solution 3 (2 minute solve, fast) by MegaMath== | ||
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https://www.youtube.com/watch?v=77UBBu1bKxk | https://www.youtube.com/watch?v=77UBBu1bKxk | ||
+ | don't recommend but its quite clean, learn what you must- Orion 2010 | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/ktzijuZtDas&t=1585 | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2024|num-b=14|num-a=16}} | {{AMC8 box|year=2024|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | [[Category:Introductory Number Theory Problems]] |
Latest revision as of 04:48, 15 June 2024
Contents
- 1 Problem 15
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3 (Answer Choices)
- 5 Video Solution 1 by Math-X (First fully understand the problem!!!)
- 6 Video Solution 2 (easy to digest) by Power Solve
- 7 Video Solution 3 (2 minute solve, fast) by MegaMath
- 8 Video Solution 4 by SpreadTheMathLove
- 9 Video Solution by NiuniuMaths (Easy to understand!)
- 10 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 11 Video Solution by Interstigation
- 12 See Also
Problem 15
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 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)
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
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 by Math-X (First fully understand the problem!!!)
https://youtu.be/BaE00H2SHQM?si=4za1LGPg_w2gwosm&t=3484
~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
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=V-xN8Njd_Lc
~NiuniuMaths
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=77UBBu1bKxk don't recommend but its quite clean, learn what you must- Orion 2010
Video Solution by Interstigation
https://youtu.be/ktzijuZtDas&t=1585
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.