Awesome Binary Operation

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0 replies

jlacosta

Mar 2, 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

Red Mop Chances

imagien_bad 32

N a minute ago by xTimmyG

What are my chances of making red mop with a 35 on jmo?

32 replies

imagien_bad

Yesterday at 8:27 PM

xTimmyG

a minute ago

9 Can I make MOP

Bigtree 19

N 6 minutes ago by cdm

My dream is to be on IMO team ik thats not going to happen b/c the kids that make it are like 6th mop quals :play_ball:. I somehow got a $208.5$ index this yr (118.5 on amc10+ 9 on AIME) i’m in 7th rn btw first year comp math also. I will grind so hard until like 30 hrs/week. I’m ok at proofs. made mc nats

19 replies

Bigtree

Mar 9, 2025

cdm

6 minutes ago

what is a moh

Soupboy0 3

N 6 minutes ago by cdm

can someone please tell me what a moh is

3 replies

Soupboy0

2 hours ago

cdm

6 minutes ago

Prove a polynomial has a nonreal root

KevinYang2.71 40

N 9 minutes ago by dno1467

Source: USAMO 2025/2

Let $n$ and $k$ be positive integers with $k<n$ . Let $P(x)$ be a polynomial of degree $n$ with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers $a_0,\,a_1,\,\ldots,\,a_k$ such that the polynomial $a_kx^k+\cdots+a_1x+a_0$ divides $P(x)$ , the product $a_0a_1\cdots a_k$ is zero. Prove that $P(x)$ has a nonreal root.

40 replies

KevinYang2.71

Mar 20, 2025

dno1467

9 minutes ago

Geometry

srnjbr 1

N an hour ago by ricarlos

in triangle abc, we know that bac=60. the circumcircle of the center i is tangent to the sides ab and ac at points e and f respectively. the midpoint of side bc is called m. if lines bi and ci intersect line ef at points p and q respectively, show that pmq is equilateral.

1 reply

srnjbr

Mar 19, 2025

ricarlos

an hour ago

A property of divisors

rightways 8

N 2 hours ago by de-Kirschbaum

Source: Kazakhstan NMO 2016, P1

Prove that one can arrange all positive divisors of any given positive integer around a circle so that for any two neighboring numbers one is divisible by another.

8 replies

rightways

Mar 17, 2016

de-Kirschbaum

2 hours ago

ortho conf DEF, radius MD, intersect ME,MF, collinear H,K,L

star-1ord 0

2 hours ago

Source: Estonia Final Round 2025 12-3

Let $ABC$ be an acute-angled triangle with $|AB|<|AC|$ . The altitudes $AD,BE$ and $CF$ intersect at $H$ . Let $M$ be the midpoint of $BC$ . Point $K$ is chosen on the extension of $EM$ beyond $M$ and point $L$ is chosen on the segment $FM$ such that $|MK|=|ML|=|MD|$ . Prove that points $K, L$ and $H$ are collinear.

a little harder version

Let $P$ be the projection of $D$ onto $HM$ . Prove that $PD$ bisects $\angle KPL$ .

0 replies

star-1ord

2 hours ago

0 replies

Funny system of equations in three variables

Tintarn 10

N 3 hours ago by Marcus_Zhang

Source: Baltic Way 2020, Problem 5

Find all real numbers $x,y,z$ so that
$\begin{align*} x^2 y + y^2 z + z^2 &= 0 \\ z^3 + z^2 y + z y^3 + x^2 y &= \frac{1}{4}(x^4 + y^4). \end{align*}$

10 replies

Tintarn

Nov 14, 2020

Marcus_Zhang

3 hours ago

a^{2m}+a^{n}+1 is perfect square

kmh1 1

N 3 hours ago by kmh1

Source: own

Find all positive integer triplets $(a,m,n)$ such that $2m>n$ and $a^{2m}+a^{n}+1$ is a perfect square.

1 reply

kmh1

Mar 20, 2025

kmh1

3 hours ago

Interesting problem

deraxenrovalo 0

4 hours ago

Given $\triangle$ $ABC$ with circumcenter $O$ $.\;$ Let $P$ be an arbitrary point on $(BOC)$ such that $P$ is outside $(ABC)$ $.\;$ Let $Q$ be an arbitrary point on $(ABC)$ $.\;$ $AB$ cuts $(ACP)$ again at $E$ and $AC$ cuts $(ABP)$ again at $F$ $.\;$ The intersection of $BF$ and $CE$ is $R$ $.\;$ Let $X$ and $Y$ be the intersection of $EF$ with $(PQC)$ and $(PQR)$ respectively such that $X$ , $Y$ , $P$ are pairwise distinct.
Show that : $(APX)$ , $(BPY)$ , $(QPE)$ are coaxial circles

hint

invert at P

0 replies

deraxenrovalo

4 hours ago

0 replies

Vieta Jumping Unsolved(Reposted)

Eagle116 0

4 hours ago

Source: MONT, Vieta Jumping part

The question is:
Let $x_1$ , $x_2$ , $\dots$ , $x_n$ be $n$ integers. If $k>n$ is an integer, prove that the only solution to
$x_1^2 + x_2^2 + \dots + x_n^2 = kx_1x_2\dots x_n$ is is $x_1 = x_2 = \dots = x_n = 0$ .

0 replies

Eagle116

4 hours ago

0 replies

Geometry with parallel lines.

falantrng 32

N 4 hours ago by endless_abyss

Source: RMM 2018,D1 P1

Let $ABCD$ be a cyclic quadrilateral an let $P$ be a point on the side $AB.$ The diagonals $AC$ meets the segments $DP$ at $Q.$ The line through $P$ parallel to $CD$ mmets the extension of the side $CB$ beyond $B$ at $K.$ The line through $Q$ parallel to $BD$ meets the extension of the side $CB$ beyond $B$ at $L.$ Prove that the circumcircles of the triangles $BKP$ and $CLQ$ are tangent .

32 replies

falantrng

Feb 24, 2018

endless_abyss

4 hours ago

sum divides n-th moment

navi_09220114 1

N 4 hours ago by ja.

Source: Own. Malaysian IMO TST 2025 P9

Given four distinct positive integers $a<b<c<d$ such that $\gcd(a,b,c,d)=1$ , find the maximum possible number of integers $1\le n\le 2025$ such that $a+b+c+d\mid a^n+b^n+c^n+d^n$
Proposed by Ivan Chan Kai Chin

1 reply

navi_09220114

Yesterday at 1:07 PM

ja.

4 hours ago

Find all functions

Jackson0423 0

4 hours ago

Find all functions F:R->R such that
1/(F(F(x))-F(x))=F(x)
I know x+1/x works..

0 replies

Jackson0423

4 hours ago

0 replies

Awesome Binary Operation

shiningsunnyday 48

N Nov 2, 2021 by megarnie

Source: 2016 AMC 10A #23

A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $p+q?$

$\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601$

48 replies

shiningsunnyday

Feb 3, 2016

megarnie

Nov 2, 2021

Awesome Binary Operation

G H J

Reveal topic tags

G H BBookmark kLocked kLocked NReply

Source: 2016 AMC 10A #23

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shiningsunnyday

1350 posts

shiningsunnyday

#1 h Feb 3, 2016, 2:26 PM • 4 Y

Y by DanielL2000, Shadow26, megarnie, Adventure10

This post has been edited 2 times. Last edited by djmathman, Feb 3, 2016, 4:12 PM

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DanielL2000

985 posts

DanielL2000

#2 h Feb 3, 2016, 2:27 PM • 6 Y

Y by WalkerTesla, spartan168, math101010, yrnsmurf, TonyStark202, Adventure10

It's just division!
So simply once you observe that you can easily solve for x to get 109.

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mishka1980

409 posts

mishka1980

#3 h Feb 3, 2016, 2:27 PM • 2 Y

Y by Harry0531, Adventure10

Just recongnize that the binary operation $\diamond$ is equal to division.

EDIT: Darn Sniped

This post has been edited 1 time. Last edited by mishka1980, Feb 3, 2016, 2:27 PM

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adihaya

1632 posts

adihaya

#4 h Feb 3, 2016, 2:27 PM • 4 Y

Y by Alnitak, elitechicken, Harry0531, Adventure10

It's division. I just somehow came up with division and this turned out to be (IMO) the easiest problem on the test. Lol

But it was hinted at pretty hard with $a\diamond a=1$ etc.

EDIT: re=sniped

This post has been edited 2 times. Last edited by adihaya, Feb 3, 2016, 2:28 PM

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Darth37

71 posts

Darth37

#5 h Feb 3, 2016, 2:28 PM • 2 Y

Y by Harry0531, Adventure10

I know, right?

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Mudkipswims42

8867 posts

Mudkipswims42

#6 h Feb 3, 2016, 2:28 PM • 2 Y

Y by mathsolver101, Adventure10

Noooo!!! Some how I didn't realize this and simplified it to $x*6*2016=100$

RIP JMO and hopefully not AIME oops wait I think I got 104

hopefully this makes AIME...

This post has been edited 3 times. Last edited by Mudkipswims42, Feb 3, 2016, 2:36 PM

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62861

3564 posts

62861

#7 h Feb 3, 2016, 2:29 PM • 6 Y

Y by AstrapiGnosis, AKAL3, Imayormaynotknowcalculus, HamstPan38825, Adventure10, Mango247

12A #20.

Quite funny problem.

Z K Y

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speck

1727 posts

speck

#8 h Feb 3, 2016, 2:31 PM • 15 Y

Y by ThisIsASentence, CornSaltButter, PiDude314, yojan_sushi, gradysocool, daniellionyang, DominicanAOPSer, Rubytown, claserken, Toinfinity, Stan2023, Adventure10, Mango247, Math4Life7, aidan0626

Proof that it's division

Plug in $(a,a,a)$ to get $a()1=a$ . Plug in $(a,b,b)$ to get $a()1=a()b \cdot b \rightarrow a = a()b \cdot b \rightarrow a()b=\dfrac{a}{b}$ .

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dhusb45

365 posts

dhusb45

#9 h Feb 3, 2016, 2:37 PM • 12 Y

Y by r3mark, acegikmoqsuwy2000, speck, checkmatetang, Mathlete55, DominicanAOPSer, LostinNowhere, shellysuper, Toinfinity, Adventure10, Mango247, Math4Life7

Proof Afterwards Without Division

$2016 \Diamond (6 \Diamond x)=(2016 \Diamond 6)*x$ , then find $(2016\Diamond6)$ :
Multiply $(2016\Diamond6)$ by 6 to get $(2016\Diamond6)*6=2016\Diamond1=2016$ , so $2016\Diamond6=336$
Then $\frac{100}{336}=\frac{25}{84}$ and $25+84=109$

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kangrui

578 posts

kangrui

#10 h Feb 3, 2016, 2:45 PM • 1 Y

Y by Adventure10

How do you know $2016\diamond 1 = 2016$ ? I got stuck there.

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dhusb45

365 posts

dhusb45

#11 h Feb 3, 2016, 2:47 PM • 2 Y

Y by Adventure10, Mango247

kangrui wrote:

How do you know $2016\diamond 1 = 2016$ ? I got stuck there.

So $2016 \Diamond 1=2016 \Diamond (2016 \Diamond 2016)=(2016 \Diamond 2016)*2016=2016$

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mathguy5041

2659 posts

mathguy5041

#12 h Feb 3, 2016, 3:23 PM • 5 Y

Y by somepersonoverhere, gradysocool, DominicanAOPSer, Adventure10, Mango247

DanielL2000 wrote:

It's just division!
So simply once you observe that you can easily solve for x to get 109.

OMG HOW DID I NOT SEE THIS!?!?!?
SO TROLL
Facepalm

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mathtastic

3258 posts

mathtastic

#13 h Feb 3, 2016, 3:25 PM • 2 Y

Y by Adventure10, Mango247

Let # be the diamond because I'm on my phone

a#b=1/b(b*a#b)=1/b(a#(b#b))=1/b(a#1)=1/b(a#a#a)=1/b(a(a#a))=a/b

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Alnitak

18 posts

Alnitak

#14 h Feb 3, 2016, 3:28 PM • 4 Y

Y by somepersonoverhere, Mudkipswims42, Adventure10, Mango247

ariedel wrote:

Noooo!!! Some how I didn't realize this and simplified it to $x*6*2016=100$

RIP JMO and hopefully not AIME oops wait I think I got 104

hopefully this makes AIME...

Just do well next time on 10B.

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samirnaqvi

12 posts

samirnaqvi

#15 h Feb 3, 2016, 3:52 PM • 2 Y

Y by Adventure10, Mango247

what question on 12 #

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sojourner1

168 posts

sojourner1

#37 h Feb 4, 2016, 4:53 AM • 6 Y

Y by pandabear10, champion999, dantx5, Imayormaynotknowcalculus, Adventure10, Mango247

I thought about this all night and couldn't find the answer... then the next day a guy I talked to was like "oh yeah it's just division" and I literally slammed my binder on the floor.
Oh well r i p I really needed this ez question for >130 AMC 12 score but whatever this was slightly comedic anyway.

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Wave-Particle

3690 posts

Wave-Particle

#38 h Feb 4, 2016, 4:54 AM • 2 Y

Y by Adventure10, Mango247

sojourner1 wrote:

Same lol

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muti66

931 posts

muti66

#39 h Feb 4, 2016, 4:55 AM • 2 Y

Y by Adventure10, Mango247

...I thought that this operation did stuff IN binary.... I thought it was going to be something like, you convert the number into binary, the square and subtract from each other and square again, or something complicated like that, so I skipped it...

This post has been edited 1 time. Last edited by muti66, Feb 4, 2016, 5:00 AM

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mathwiz_aku

1001 posts

mathwiz_aku

#40 h Feb 4, 2016, 4:58 AM • 2 Y

Y by Adventure10, Mango247

You know... we could make some many excuses for not getting this problem right...

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adihaya

1632 posts

adihaya

#41 h Feb 4, 2016, 5:09 AM • 2 Y

Y by Adventure10, Mango247

This was my favorite problem on the test. If you solved it during the test, you would feel a great ecstasy—it was like the MAA was giving you a hug and apologizing for last year's #14 (note

though a true apology would be an average-difficulty test *cough cough maa*

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62861

3564 posts

62861

#42 h Feb 4, 2016, 5:11 AM • 2 Y

Y by Adventure10, Mango247

adihaya wrote:

This was my favorite problem on the test. If you solved it during the test, you would feel a great ecstasy—it was like the MAA was giving you a hug and apologizing for last year's #14 (note

though a true apology would be an average-difficulty test *cough cough maa*

I wouldn't say ecstasy. I was just happy I got a free problem on 12A.

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Snowie

275 posts

Snowie

#43 h Feb 4, 2016, 1:05 PM • 2 Y

Y by Adventure10, Mango247

I absolutely hated this problem during the test... I read it 10 times and I got more confused after each one. And I solved wrong at the end as well

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mathwizard888

1635 posts

mathwizard888

#44 h Feb 4, 2016, 2:16 PM • 2 Y

Y by Adventure10, Mango247

adihaya wrote:

This was my favorite problem on the test. If you solved it during the test, you would feel a great ecstasy—it was like the MAA was giving you a hug and apologizing for last year's #14 (note

though a true apology would be an average-difficulty test *cough cough maa*

This test was average difficulty...

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v_Enhance

6869 posts

v_Enhance

#45 h Jul 3, 2019, 6:53 AM • 25 Y

Y by khina, cosmicgenius, Mudkipswims42, bangzheng, sub_math, yrnsmurf, sunfishho, mathisawesome2169, Hexagrammum16, Toinfinity, Pendronator, mathleticguyyy, Synapse2718, Imayormaynotknowcalculus, coolmath_2018, fjm30, Aryan-23, v4913, IAmTheHazard, HamstPan38825, tigerzhang, vsamc, ThinGarfield, CoolCarsOnTheRun, Adventure10

Some problems grow up.

This post has been edited 1 time. Last edited by v_Enhance, Jul 3, 2019, 6:54 AM

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Pendronator

51 posts

Pendronator

#49 h Dec 30, 2019, 8:39 PM • 1 Y

Y by Adventure10

I got 109.

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SharpApricot123

1270 posts

SharpApricot123

#50 h Aug 9, 2020, 12:40 AM • 1 Y

Y by Aryan-23

solution

After we observe that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and $a\,\diamondsuit \,a=1$
means that the binary operation $\diamondsuit$ can be substituted with the division symbol, we can solve this problem.
$2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100\Longrightarrow 2016 \div (6\div x)=2016\div\frac{6}{x}=100$ This equals : $\frac{2016x}{6}=100$ We multiply both sides by $6$ , $2016x=600\Longrightarrow x=\frac{600}{2016}=\frac{150}{504}=\frac{25}{84}$ $84+25=\boxed{109}$

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Pleaseletmewin

1574 posts

Pleaseletmewin

#51 h Nov 6, 2020, 5:38 PM

Y by

I can't believe I didn't see this when mocking the test. Note that $\diamondsuit$ is just division. Hence, all we have to do is $\frac{2016}{\frac{6}{x}}=100$ which yields $x=\frac{25}{84}$ which means the answer is $109$ .

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ilovepizza2020

12156 posts

ilovepizza2020

#52 h Nov 6, 2020, 5:43 PM

Y by

Pleaseletmewin wrote:

Same I somehow just missed it

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kante314

3099 posts

kante314

#53 h Sep 7, 2021, 2:57 AM • 2 Y

Y by mahaler, Geometry285

Notice that
$(a \hspace{1mm} \diamondsuit \hspace{1mm} b) \cdot b = a \hspace{1mm} \diamondsuit \hspace{1mm} (b \hspace{1mm} \diamondsuit \hspace{1mm} b) = a \hspace{1mm} \diamondsuit \hspace{1mm} 1 = a \hspace{1mm} \diamondsuit \hspace{1mm} ( a \hspace{1mm} \diamondsuit \hspace{1mm} a) = ( a \hspace{1mm} \diamondsuit \hspace{1mm} a) \cdot a = a$ Thus, $(a \hspace{1mm} \diamondsuit \hspace{1mm} b)=\frac{a}{b}$ So, $2016 \hspace{1mm} \diamondsuit \hspace{1mm} (6 \hspace{1mm} \diamondsuit \hspace{1mm} x)=100 \implies (2016 \hspace{1mm} \diamondsuit \hspace{1mm} 6) \cdot x=100 \implies 336x=100 \implies x=\frac{25}{84}$ Therefore, the answer is $25+84=\boxed{109}$

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megarnie

5538 posts

megarnie

#54 h Nov 2, 2021, 5:05 PM

Y by

Solution

Let $P(a,b,c)$ denote the given assertion.

Claim: $\diamondsuit$ is "injective" (if $a\diamondsuit b =a\diamondsuit c$ , then $b=c$ ).
Proof:
$P(a,a,b): a\diamondsuit(a\diamondsuit b)=(a\diamondsuit a)\cdot b$ .
$P(a,a,c): a\diamondsuit(a\diamondsuit b)=(a\diamondsuit a)\cdot c$ .

Since $a\diamondsuit a\ne0$ , $b=c$ .

$P(a,a,1): a\diamondsuit(a\diamondsuit 1)=a\diamondsuit a$ , so $a\diamondsuit 1=a$ .

$P(a,1,b): a\diamondsuit (1\diamondsuit b)=a\cdot b$ .

$P(1,1,b): 1\diamondsuit (1\diamondsuit b)=b$ .

$P(a,1, 1\diamondsuit b): a\diamondsuit b=a\cdot (1\diamondsuit b)$ .

Set $f(x)=1\diamondsuit x$ . So $1=a\cdot f(a)\implies f(a)=\frac{1}{a}$ .

So $a\diamondsuit b=a\cdot f(b)=\frac{a}{b}$ , which clearly works.

Thus $2016\diamondsuit (\frac{6}{x})=\frac{2016x}{6}=336x=100$ , so $x=\frac{25}{84}\implies \boxed{A}$ .

This post has been edited 1 time. Last edited by megarnie, Nov 2, 2021, 5:05 PM

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