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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
USAMO MERCH
elasticwealth   1
N 2 minutes ago by zhoujef000
Jane street sent:
- A T-shirt
- Deck of cards
- Portable charger (battery)
- 12 oz mug
- Jane Street Hat
1 reply
elasticwealth
9 minutes ago
zhoujef000
2 minutes ago
USAMO Medals
YauYauFilter   21
N 24 minutes ago by hashbrown2009
YauYauFilter
Apr 24, 2025
hashbrown2009
24 minutes ago
Is EGMO good for JMO Geometry Questions?
MathRook7817   4
N 27 minutes ago by MathRook7817
Hi guys, I was just wondering if EGMO is a good book for JMO/AMO/olympiad level questions, or if there exists another olympiad geo book. Thanks!
4 replies
1 viewing
MathRook7817
Today at 3:05 AM
MathRook7817
27 minutes ago
Incentre-excentre geometry
oVlad   1
N an hour ago by mashumaro
Source: Romania Junior TST 2025 Day 2 P2
Consider a scalene triangle $ABC$ with incentre $I$ and excentres $I_a,I_b,$ and $I_c$, opposite the vertices $A,B,$ and $C$ respectively. The incircle touches $BC,CA,$ and $AB$ at $E,F,$ and $G$ respectively. Prove that the circles $IEI_a,IFI_b,$ and $IGI_c$ have a common point other than $I$.
1 reply
oVlad
2 hours ago
mashumaro
an hour ago
Two equal angles
jayme   5
N an hour ago by Captainscrubz
Dear Mathlinkers,

1. ABCD a square
2. I the midpoint of AB
3. 1 the circle center at A passing through B
4. Q the point of intersection of 1 with the segment IC
5. X the foot of the perpendicular to BC from Q
6. Y the point of intersection of 1 with the segment AX
7. M the point of intersection of CY and AB.

Prove : <ACI = <IYM.

Sincerely
Jean-Louis
5 replies
jayme
May 2, 2025
Captainscrubz
an hour ago
Geometry
Lukariman   1
N an hour ago by Lukariman
Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that <HDM = 2∠AMP.
1 reply
Lukariman
2 hours ago
Lukariman
an hour ago
Inequality involving square root cube root and 8th root
bamboozled   3
N 2 hours ago by Jackson0423
If $a,b,c,d,e,f,g,h,k\in R^+$ and $a+b+c=d+e+f=g+h+k=8$, then find the maximum value of $\sqrt{ad^3 g^4} +\sqrt[3]{be^3 h^4} + \sqrt[8]{cf^3 k^4}$
3 replies
bamboozled
Today at 4:46 AM
Jackson0423
2 hours ago
Geo metry
TUAN2k8   1
N 3 hours ago by SimplisticFormulas
Help me plss!
Given an acute triangle $ABC$. Points $D$ and $E$ lie on segments $AB$ and $AC$, respectively. Lines $BD$ and $CE$ intersect at point $F$. The circumcircles of triangles $BDF$ and $CEF$ intersect at a second point $P$. The circumcircles of triangles $ABC$ and $ADE$ intersect at a second point $Q$. Point $K$ lies on segment $AP$ such that $KQ \perp AQ$. Prove that triangles $\triangle BKD$ and $\triangle CKE$ are similar.
1 reply
TUAN2k8
4 hours ago
SimplisticFormulas
3 hours ago
Nordic 2025 P3
anirbanbz   9
N Today at 8:11 AM by Tsikaloudakis
Source: Nordic 2025
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. Let $E$ and $F$ be points on the line segments $AC$ and $AB$ respectively such that $AEHF$ is a parallelogram. Prove that $\vert OE \vert = \vert OF \vert$.
9 replies
anirbanbz
Mar 25, 2025
Tsikaloudakis
Today at 8:11 AM
Aime type Geo
ehuseyinyigit   1
N Today at 7:44 AM by ehuseyinyigit
Source: Turkish First Round 2024
In a scalene triangle $ABC$, let $M$ be the midpoint of side $BC$. Let the line perpendicular to $AC$ at point $C$ intersect $AM$ at $N$. If $(BMN)$ is tangent to $AB$ at $B$, find $AB/MA$.
1 reply
ehuseyinyigit
Yesterday at 9:04 PM
ehuseyinyigit
Today at 7:44 AM
Arbitrary point on BC and its relation with orthocenter
falantrng   34
N Today at 7:41 AM by Mamadi
Source: Balkan MO 2025 P2
In an acute-angled triangle \(ABC\), \(H\) be the orthocenter of it and \(D\) be any point on the side \(BC\). The points \(E, F\) are on the segments \(AB, AC\), respectively, such that the points \(A, B, D, F\) and \(A, C, D, E\) are cyclic. The segments \(BF\) and \(CE\) intersect at \(P.\) \(L\) is a point on \(HA\) such that \(LC\) is tangent to the circumcircle of triangle \(PBC\) at \(C.\) \(BH\) and \(CP\) intersect at \(X\). Prove that the points \(D, X, \) and \(L\) lie on the same line.

Proposed by Theoklitos Parayiou, Cyprus
34 replies
falantrng
Apr 27, 2025
Mamadi
Today at 7:41 AM
Parallelograms and concyclicity
Lukaluce   31
N Today at 4:15 AM by Ihatecombin
Source: EGMO 2025 P4
Let $ABC$ be an acute triangle with incentre $I$ and $AB \neq AC$. Let lines $BI$ and $CI$ intersect the circumcircle of $ABC$ at $P \neq B$ and $Q \neq C$, respectively. Consider points $R$ and $S$ such that $AQRB$ and $ACSP$ are parallelograms (with $AQ \parallel RB, AB \parallel QR, AC \parallel SP$, and $AP \parallel CS$). Let $T$ be the point of intersection of lines $RB$ and $SC$. Prove that points $R, S, T$, and $I$ are concyclic.
31 replies
Lukaluce
Apr 14, 2025
Ihatecombin
Today at 4:15 AM
purple comet discussion
ConfidentKoala4   64
N Today at 3:11 AM by MathCosine
when can we discuss purple comet
64 replies
2 viewing
ConfidentKoala4
May 2, 2025
MathCosine
Today at 3:11 AM
Concurrency from isogonal Mittenpunkt configuration
MarkBcc168   17
N Today at 12:25 AM by Ilikeminecraft
Source: Fake USAMO 2020 P3
Let $\triangle ABC$ be a scalene triangle with circumcenter $O$, incenter $I$, and incircle $\omega$. Let $\omega$ touch the sides $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$ at points $D$, $E$, and $F$ respectively. Let $T$ be the projection of $D$ to $\overline{EF}$. The line $AT$ intersects the circumcircle of $\triangle ABC$ again at point $X\ne A$. The circumcircles of $\triangle AEX$ and $\triangle AFX$ intersect $\omega$ again at points $P\ne E$ and $Q\ne F$ respectively. Prove that the lines $EQ$, $FP$, and $OI$ are concurrent.

Proposed by MarkBcc168.
17 replies
MarkBcc168
Apr 28, 2020
Ilikeminecraft
Today at 12:25 AM
FLYFLYFLYBUGBUGBUG
rnatog337   32
N Apr 25, 2025 by Ilikeminecraft
Source: 2024 AMC 8 P15
Let the letters $F$, $L$, $Y$, $B$, $U$, $G$ represent different digits. Suppose $\underline{F}\underline{L}\underline{Y}\underline{F}\underline{L}\underline{Y}$ is the largest number that satisfies the equation $$8 \cdot \underline{F}\underline{L}\underline{Y}\underline{F}\underline{L}\underline{Y} = \underline{B}\underline{U}\underline{G}\underline{B}\underline{U}\underline{G}.$$What is the value of $\underline{F}\underline{L}\underline{Y} + \underline{B}\underline{U}\underline{G}$?

$\textbf{(A) } 1089\qquad\textbf{(B) } 1098\qquad\textbf{(C) } 1107\qquad\textbf{(D) } 1116\qquad\textbf{(E) } 1125$
32 replies
rnatog337
Jan 25, 2024
Ilikeminecraft
Apr 25, 2025
FLYFLYFLYBUGBUGBUG
G H J
G H BBookmark kLocked kLocked NReply
Source: 2024 AMC 8 P15
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rnatog337
410 posts
#1 • 2 Y
Y by megarnie, Rounak_iitr
Let the letters $F$, $L$, $Y$, $B$, $U$, $G$ represent different digits. Suppose $\underline{F}\underline{L}\underline{Y}\underline{F}\underline{L}\underline{Y}$ is the largest number that satisfies the equation $$8 \cdot \underline{F}\underline{L}\underline{Y}\underline{F}\underline{L}\underline{Y} = \underline{B}\underline{U}\underline{G}\underline{B}\underline{U}\underline{G}.$$What is the value of $\underline{F}\underline{L}\underline{Y} + \underline{B}\underline{U}\underline{G}$?

$\textbf{(A) } 1089\qquad\textbf{(B) } 1098\qquad\textbf{(C) } 1107\qquad\textbf{(D) } 1116\qquad\textbf{(E) } 1125$
This post has been edited 2 times. Last edited by rnatog337, Jan 25, 2024, 5:45 PM
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zhoujef000
315 posts
#2 • 5 Y
Y by littlefox_amc, weihang, Spiritpalm, ESAOPS, megarnie
lol I put D since i didnt read distinct oops
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rnatog337
410 posts
#3 • 1 Y
Y by megarnie
I'm pretty sure the configuration $\underline{FLY} = 123$ works. Then, we get C.
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sundavid2591
114 posts
#4 • 1 Y
Y by megarnie
Rewrite this as 8 FLY = BUG by dividing both sides by 1001. Note that BUG must be a 3 digit number and is divisible by 8. Testing values starting from 1000 as BUG should be as large as possible, we find that 992 doesn't satisfy the conditions, but 984 does. It follows that FLY = 123, and 9 * FLY = 1107
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baassid24
196 posts
#5 • 1 Y
Y by megarnie
Here is my solution:
Click to reveal hidden text
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Countmath1
180 posts
#6 • 1 Y
Y by megarnie
We must have FLY < 124999 = floor(999999/8) so the two best choices are 124 (doesnt work) and 123(yields BUG = 984) so the ans is $123 + 984 = \boxed{\textbf{(C)}\ 1107}.$
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DuoDuoling0
3859 posts
#7 • 2 Y
Y by megarnie, axsolers_24
There is symmetry in this problem :)

Solution
This post has been edited 1 time. Last edited by DuoDuoling0, Jan 26, 2024, 1:35 AM
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ihatemath123
3446 posts
#8 • 1 Y
Y by megarnie
Unironically a very nice AMC 8 problem :coolspeak: :coolspeak:

We must have $8 \cdot FLY = BUG$, so $FLY \leq 124$. However, $124$ results in $B=U=9$, which is not allowed, thus $FLY \leq 123$. In fact, $FLY = 123$ results in $BUG = 984$ which is allowed. Our final answer is $123 + 984 = \boxed{\textbf{(C) } 1107}$.
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Yrock
1289 posts
#9 • 1 Y
Y by megarnie
(C) because I tried out 123, then 124 and 125 and they all fail
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aidan0626
1897 posts
#10 • 1 Y
Y by megarnie
luckily i noticed it said distinct
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YIYI-JP
2175 posts
#11 • 2 Y
Y by matchaisgreen, Mathusiast2012
I hated this problem, I was stuck and I just guessed
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Elephant200
1472 posts
#12
Y by
Solution

I sillied the problem though (forgot 992 is invalid :( )
This post has been edited 3 times. Last edited by Elephant200, Jan 26, 2024, 2:57 PM
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yambe2002
1725 posts
#13
Y by
i put D for some reason when i got the same sort of problem correct while practicing??
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megahertz13
3183 posts
#14
Y by
Two minute solution (fast)
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MathWizard10
1434 posts
#15
Y by
i used the answer choices to solve it
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efshipaio
13 posts
#16
Y by
C maybe because trivially if we think,we will take FLY=123
BUG=984
Sum- C
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happypi31415
747 posts
#17
Y by
rip i got trolled by the 'distinct'
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britishprobe17
211 posts
#18
Y by
megahertz13 wrote:
Two minute solution (fast)

speedrunner AMC 8 2024 type
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niuaccount
349 posts
#19
Y by
happypi31415 wrote:
rip i got trolled by the 'distinct'

same
i feel like 50% of people got trolled on that point
Z K Y
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Catcumber
162 posts
#20
Y by
niuaccount wrote:
happypi31415 wrote:
rip i got trolled by the 'distinct'

same
i feel like 50% of people got trolled on that point



more like 80%
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dbnl
3377 posts
#21
Y by
atp i get trolled by these keywords so many times it's just a matter of luck if i see them lmho like if it's a good day I'll read the problem carefully :yup:
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Gavin_Deng
808 posts
#22
Y by
zhoujef000 wrote:
lol I put D since i didnt read distinct oops

Me too
Z K Y
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Catcumber
162 posts
#23
Y by
Gavin_Deng wrote:
zhoujef000 wrote:
lol I put D since i didnt read distinct oops

Me too

same here!
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LostDreams
144 posts
#24
Y by
skibidi problem

The largest that $F$ can be is $1$ so let $F$ equal to $1$ since $F$ can't be greater than $2$ otherwise it would become a $7$ digit number, which wouldn't satisfy the expression to the right of the equation.

By the idea that the expression to the right can't be a $7$ digit number we note that $L$ can either be $0$ or $2$ as the digits must be distinct since $L<3$.

Looking through all the cases from greatest to least and noting that all the digits must be distinct we find that $FLY = 123$ hence $BUG = 984$.

Giving us $984+123=\boxed{\textbf{(C) } 1107}$
This post has been edited 1 time. Last edited by LostDreams, Mar 6, 2024, 4:33 PM
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GreenMagnify
4 posts
#26
Y by
Bonjour! Comont Ca Va
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reasryra
178 posts
#28
Y by
zhoujef000 wrote:
lol I put D since i didnt read distinct oops

I did the same thing. I got the right answer because I read distinct and did it right. Came back to it and was like hey this value is better when it wasn't distinct. LOL!

EDIT: Ended up getting it wrong.
This post has been edited 1 time. Last edited by reasryra, Mar 25, 2024, 8:21 PM
Reason: gg
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JH_K2IMO
126 posts
#29
Y by
8 x (FLYFLY)=(BUGBUG).
FLYFLY =FLY x 1001, BUGBUG =BUG x 1001.
8 x FLY x 1001=BUG x 1001.
Therefore, 8x FLY=BUG.
100<= FLY<=124.
F, L, Y cannot be the same.
∴FLY=123, BUG=123 x 8=984.
FLY+BUG=123+984=1107.
The answer is (C)1107.
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xTimmyG
265 posts
#30
Y by
there are a lot of repeating numbers, so it has to be C,D, or E. Also, since A and B have 8 and 9, we expect the answer to have a 7. Thus, the answer is C. trivial amc8 problem.
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SpeedCuber7
1833 posts
#31
Y by
xTimmyG wrote:
there are a lot of repeating numbers, so it has to be C,D, or E. Also, since A and B have 8 and 9, we expect the answer to have a 7. Thus, the answer is C. trivial amc8 problem.

rbo calls the hardest problem on the test trivial
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Challengees24
1098 posts
#32 • 1 Y
Y by Pengu14
SpeedCuber7 wrote:
xTimmyG wrote:
there are a lot of repeating numbers, so it has to be C,D, or E. Also, since A and B have 8 and 9, we expect the answer to have a 7. Thus, the answer is C. trivial amc8 problem.

rbo calls the hardest problem on the test trivial

hardest problem on test :skull:
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Challengees24
1098 posts
#33
Y by
xTimmyG wrote:
there are a lot of repeating numbers, so it has to be C,D, or E. Also, since A and B have 8 and 9, we expect the answer to have a 7. Thus, the answer is C. trivial amc8 problem.

this is a horrible method
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sadas123
1261 posts
#34
Y by
rnatog337 wrote:
Let the letters $F$, $L$, $Y$, $B$, $U$, $G$ represent different digits. Suppose $\underline{F}\underline{L}\underline{Y}\underline{F}\underline{L}\underline{Y}$ is the largest number that satisfies the equation $$8 \cdot \underline{F}\underline{L}\underline{Y}\underline{F}\underline{L}\underline{Y} = \underline{B}\underline{U}\underline{G}\underline{B}\underline{U}\underline{G}.$$What is the value of $\underline{F}\underline{L}\underline{Y} + \underline{B}\underline{U}\underline{G}$?

$\textbf{(A) } 1089\qquad\textbf{(B) } 1098\qquad\textbf{(C) } 1107\qquad\textbf{(D) } 1116\qquad\textbf{(E) } 1125$

Pretty easy problem here is sol, We can find the greatest FLYFLY when multiplied by 8 it is still 6 digits we get 123 and when we multiply 123123x8 we get 984984 then we get 123+984 which gives us the answer of 1107! I sillied on this question on the actual AMC 8 because I was in 5th grade but know I finally solved it.
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Ilikeminecraft
616 posts
#36
Y by
Solved with ST2009 and Awesomeness_in_a_bun

fly is 123 works and everything higher fails
thus, the answer is $123 \cdot 9 = 1107$ which is C
This post has been edited 1 time. Last edited by Ilikeminecraft, Apr 25, 2025, 6:15 PM
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