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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.

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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
Calculus
youochange   14
N 11 minutes ago by Moubinool
Find the area enclosed by the curves $e^x,e^{-x},x^2+y^2=1$
14 replies
youochange
Yesterday at 2:38 PM
Moubinool
11 minutes ago
Interesting inequalities
sqing   2
N 26 minutes ago by sqing
Source: Own
Let $ a,b>0 $ and $ a+b\leq 1  $ . Prove that
$$\left(\frac{4}{a^2}-1\right)\left(\frac{4}{b^2}-1\right)-k\left(\frac{a}{b}+\frac{b}{a}\right) \geq 225-2k$$Where $104\geq k\in N^+.$
$$\left(\frac{4}{a^2}-1\right)\left(\frac{4}{b^2}-1\right) \geq 225 $$$$\left(\frac{4}{a^2}-1\right)\left(\frac{4}{b^2}-1\right)-106\left(\frac{a}{b}+\frac{b}{a}\right) \geq \frac{155}{12}$$$$\left(\frac{4}{a^2}-1\right)\left(\frac{4}{b^2}-1\right)-108\left(\frac{a}{b}+\frac{b}{a}\right) \geq \frac{26}{3}$$$$\left(\frac{4}{a^2}-1\right)\left(\frac{4}{b^2}-1\right)-110\left(\frac{a}{b}+\frac{b}{a}\right) \geq \frac{17}{4}$$
2 replies
1 viewing
sqing
an hour ago
sqing
26 minutes ago
Combi Geo
Adywastaken   3
N 28 minutes ago by TUAN2k8
Source: NMTC 2024/8
A regular polygon with $100$ vertices is given. To each vertex, a natural number from the set $\{1,2,\dots,49\}$ is assigned. Prove that there are $4$ vertices $A, B, C, D$ such that if the numbers $a, b, c, d$ are assigned to them respectively, then $a+b=c+d$ and $ABCD$ is a parallelogram.
3 replies
Adywastaken
Yesterday at 3:58 PM
TUAN2k8
28 minutes ago
The familiar right angle from the orthocenter
buratinogigle   3
N 34 minutes ago by luutrongphuc
Source: Own, HSGSO P6
Let $ABC$ be a triangle inscribed in a circle $\omega$ with orthocenter $H$ and altitude $BE$. Let $M$ be the midpoint of $AH$. Line $BM$ meets $\omega$ again at $P$. Line $PE$ meets $\omega$ again at $Q$. Let $K$ be the orthogonal projection of $E$ on the line $BC$. Line $QK$ meets $\omega$ again at $G$. Prove that $GA\perp GH$.
3 replies
buratinogigle
5 hours ago
luutrongphuc
34 minutes ago
cyc sum (a+1)\sqrt{2a(1-a)} \geq 8(ab+bc+ca)
Amir Hossein   12
N 37 minutes ago by AylyGayypow009
Source: Greece JBMO TST 2017, Problem 1
Positive real numbers $a,b,c$ satisfy $a+b+c=1$. Prove that
$$(a+1)\sqrt{2a(1-a)}  + (b+1)\sqrt{2b(1-b)}  + (c+1)\sqrt{2c(1-c)}  \geq 8(ab+bc+ca).$$Also, find the values of $a,b,c$ for which the equality happens.
12 replies
Amir Hossein
Jun 25, 2018
AylyGayypow009
37 minutes ago
R to R, with x+f(xy)=f(1+f(y))x
NicoN9   2
N an hour ago by dangerousliri
Source: Own.
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that\[
x+f(xy)=f(1+f(y))x
\]for all $x, y\in \mathbb{R}$.
2 replies
NicoN9
an hour ago
dangerousliri
an hour ago
Brilliant guessing game on triples
Assassino9931   1
N 2 hours ago by Sardor_lil
Source: Al-Khwarizmi Junior International Olympiad 2025 P8
There are $100$ cards on a table, flipped face down. Madina knows that on each card a single number is written and that the numbers are different integers from $1$ to $100$. In a move, Madina is allowed to choose any $3$ cards, and she is told a number that is written on one of the chosen cards, but not which specific card it is on. After several moves, Madina must determine the written numbers on as many cards as possible. What is the maximum number of cards Madina can ensure to determine?

Shubin Yakov, Russia
1 reply
Assassino9931
Yesterday at 9:46 AM
Sardor_lil
2 hours ago
in n^2-9 has 6 positive divisors than GCD (n-3, n+3)=1
parmenides51   7
N 2 hours ago by AylyGayypow009
Source: Greece JBMO TST 2016 p3
Positive integer $n$ is such that number $n^2-9$ has exactly $6$ positive divisors. Prove that GCD $(n-3, n+3)=1$
7 replies
parmenides51
Apr 29, 2019
AylyGayypow009
2 hours ago
a deep thinking topic. either useless or extraordinary , not yet disovered
jainam_luniya   5
N 2 hours ago by jainam_luniya
Source: 1.99999999999....................................................................1. it this possible or not we can debate
it can be a new discovery in world or NT
5 replies
jainam_luniya
2 hours ago
jainam_luniya
2 hours ago
Divisibilty...
Sadigly   4
N 2 hours ago by jainam_luniya
Source: Azerbaijan Junior NMO 2025 P2
Find all $4$ consecutive even numbers, such that the sum of their squares divides the square of their product.
4 replies
Sadigly
Yesterday at 9:07 PM
jainam_luniya
2 hours ago
ioqm to imo journey
jainam_luniya   2
N 2 hours ago by jainam_luniya
only imginative ones are alloud .all country and classes or even colleges
2 replies
jainam_luniya
3 hours ago
jainam_luniya
2 hours ago
Inequality
Sadigly   5
N 2 hours ago by jainam_luniya
Source: Azerbaijan Junior MO 2025 P5
For positive real numbers $x;y;z$ satisfying $0<x,y,z<2$, find the biggest value the following equation could acquire:


$$(2x-yz)(2y-zx)(2z-xy)$$
5 replies
Sadigly
May 9, 2025
jainam_luniya
2 hours ago
D'B, E'C and l are congruence.
cronus119   7
N 3 hours ago by Tkn
Source: 2022 Iran second round mathematical Olympiad P1
Let $E$ and $F$ on $AC$ and $AB$ respectively in $\triangle ABC$ such that $DE || BC$ then draw line $l$ through $A$ such that $l || BC$ let $D'$ and $E'$ reflection of $D$ and $E$ to $l$ respectively prove that $D'B, E'C$ and $l$ are congruence.
7 replies
cronus119
May 22, 2022
Tkn
3 hours ago
a set of $9$ distinct integers
N.T.TUAN   17
N 3 hours ago by hlminh
Source: APMO 2007
Let $S$ be a set of $9$ distinct integers all of whose prime factors are at most $3.$ Prove that $S$ contains $3$ distinct integers such that their product is a perfect cube.
17 replies
N.T.TUAN
Mar 31, 2007
hlminh
3 hours ago
Disjoint Pairs
MithsApprentice   42
N Yesterday at 8:07 AM by endless_abyss
Source: USAMO 1998
Suppose that the set $\{1,2,\cdots, 1998\}$ has been partitioned into disjoint pairs $\{a_i,b_i\}$ ($1\leq i\leq 999$) so that for all $i$, $|a_i-b_i|$ equals $1$ or $6$. Prove that the sum \[ |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}|  \] ends in the digit $9$.
42 replies
MithsApprentice
Oct 9, 2005
endless_abyss
Yesterday at 8:07 AM
Disjoint Pairs
G H J
Source: USAMO 1998
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MithsApprentice
2390 posts
#1 • 7 Y
Y by nmd27082001, mathematicsy, Adventure10, megarnie, HWenslawski, Mango247, ItsBesi
Suppose that the set $\{1,2,\cdots, 1998\}$ has been partitioned into disjoint pairs $\{a_i,b_i\}$ ($1\leq i\leq 999$) so that for all $i$, $|a_i-b_i|$ equals $1$ or $6$. Prove that the sum \[ |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}|  \] ends in the digit $9$.
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paladin8
3237 posts
#2 • 5 Y
Y by vsathiam, Richangles, Adventure10, HWenslawski, Mango247
WLOG, we may assume $a_i > b_i$ or we can just switch $a_i$ and $b_i$.

So the sum just becomes $\sum a_i - \sum b_i$.

But we know $\sum a_i + \sum b_i = \frac{1998 \cdot 1999}{2} = 999 \cdot 1999$ is odd.

So $\sum a_i - \sum b_i$ is odd as well.

Since $\sum a_i - \sum b_i$ is the sum of $999$ $1$'s or $6$'s and is odd, we know there must be an even number of $6$'s. Let $2x$ be the number of $6$'s.

Then $\sum a_i - \sum b_i = 6(2x)+1(999-2x) = 10x + 999 \equiv 9 \pmod{10}$ as desired. QED.
This post has been edited 1 time. Last edited by paladin8, Dec 18, 2005, 10:14 PM
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zabelman
1072 posts
#3 • 1 Y
Y by Adventure10
Assume WLOG that $6=b_1-a_1=\cdots=b_k-a_k$ and $1=b_{k+1}-a_{k+1}=\cdots=b_n-a_n$. For $i,j\le k$, we say that $\{a_j,b_j\}$ links onto $\{a_i,b_i\}$ if $\{a_i,b_i\}$ if $a_i<a_j<b_i<b_j$ or $a_j<a_i<b_j<b_i$.

Since pair $\{a_i,b_i\}$ links onto$\{a_j,b_j\}$ if and only if $\{a_j,b_j\}$ links onto $\{a_i,b_i\}$, there are an even number of ordered pairs of integers $(i,j)$ such that pair $j$ links onto pair $i$.

For a given $i\le k$, there are an odd number of integers between $a_i$ and $b_i$, so the number of pairs that link onto$\{a_i,b_i\}$ is odd. Thus, the total number of ordered pairs of integers $(i,j)$ such that pair $j$ links onto pair $i$ is congruent to
\[ \sum_{i=1}^k(\text{number of pairs that link onto pair \textit{i}}) \equiv \sum_{i=1}^k 1\equiv k \bmod 2.  \] Therefore, since we already know that this number is even, we find that $k$ is even, and the value of the given sum is
\[ 6k+(999-k) = 999+5k\equiv 9\bmod10.  \]
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Aneo.
1111 posts
#4 • 2 Y
Y by Adventure10, Mango247
Take the equation modulo 5. Then the sum is $ 999\cdot 1\equiv4\pmod{5}$. As shown above, the equation is odd, so the sum must be 9 mod 10.
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Arvind_sn
524 posts
#5 • 2 Y
Y by Adventure10, Mango247
Here is one proof. I do not think it's very well written, especially the lemma (did I even need a lemma??) but here it is. (Sorry, I also didn't know how to make the congruence sign, and I was in a hurry, so I just put equal signs.)

lemma

main proof
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thecmd999
2860 posts
#6 • 2 Y
Y by Adventure10, Mango247
Solution
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codyj
723 posts
#7 • 3 Y
Y by vsathiam, Adventure10, Mango247
Solution by codyj and ithinksomuch
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AnonymousBunny
339 posts
#8 • 1 Y
Y by Adventure10
Solution
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Wave-Particle
3690 posts
#9 • 1 Y
Y by Adventure10
Solution
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vsathiam
201 posts
#10 • 2 Y
Y by Delray, Adventure10
Ok my sol was overkill:
Here are the two lemmas. First has been proved by many of above solutions, second is true by construction. (just need to describe it nicely.")

proof sketch
This post has been edited 1 time. Last edited by vsathiam, Aug 26, 2017, 1:13 PM
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Delray
348 posts
#11 • 2 Y
Y by vsathiam, Adventure10
Let $k$ of the differences be six, and let the rest be one. Let $A$, $B$, and $C$ denote the number of even-even, even-odd, and odd-odd pairs. Clearly, there are $B+2A=999$ evens, hence $B$ is odd. It follows that $k$ is even, so $5k+999\equiv 9 \mod 10$ as desired. $\square$
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star32
165 posts
#12 • 2 Y
Y by Mathematicsislovely, Mango247
Solution
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v_Enhance
6877 posts
#13 • 7 Y
Y by HamstPan38825, Lcz, son7, mathematicsy, ike.chen, math31415926535, Mango247
Let $S$ be the sum. Modulo $2$, \[ S = \sum |a_i-b_i| \equiv \sum (a_i+b_i) 	= 1 + 2 + \dots + 1998 \equiv 1 \pmod 2. \]Modulo $5$, \[ S = \sum |a_i-b_i| = 1 \cdot 999 \equiv 4 \pmod 5. \]So $S \equiv 9 \pmod{10}$.
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SenorIncongito
307 posts
#14
Y by
I had the same solution as @above, but I couldn't get the $1$ modulo $2$ part. Why can we add the numbers like that? (I don't really get the $1+2+...+1998$ part).
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GeronimoStilton
1521 posts
#15
Y by
@above It's because $x\equiv -x\pmod{2}$ for all integers $x$, so $|a_i-b_i|\equiv a_i-b_i\pmod{2}$ regardless of whether $a_i\ge b_i$.
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HamstPan38825
8863 posts
#16
Y by
Mod 5, the expression is obviously 4 mod 5. Now observe that $$|a_1-b_1| \equiv a_1+b_1 \pmod 2,$$implying the desired sum is congruent to $$\sum_{i=1}^{1998} i \equiv 1 \pmod 2.$$Combining gives the result.
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asdf334
7585 posts
#17 • 3 Y
Y by Math4Life2020, channing421, math31415926535
WLOG let $a_i>b_i$ and let $a_i-b_i=d_i$. Then $d_i$ equals $1$ or $6$, and we only need to show that $$\sum_{i=1}^{999}d_i\equiv 1\pmod 2.$$Notice that $$\sum_{i=1}^{999}d_i+2\sum_{i=1}^{999}b_i=\sum_{i=1}^{999}a_i+\sum_{i=1}^{999}b_i\equiv 1\pmod 2,$$and we are done. $\blacksquare$
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Tafi_ak
309 posts
#18
Y by
The number $6$ comes 'even-even' or 'odd-odd'. We claim that the number of $6$'s must be even. If odd then we get greater than $999$ odd or even number, this is a contradiction because $\{1,2,\cdots 1998\}$ have $999$ even and $999$ odd. So
\begin{eqnarray*}
\sum _{i=1}^{999} |a_i-b_i|=6\times 2k+1\times(999-2k)\equiv 10k+999\equiv 9 \hspace{6mm} \text{(mod 10)}
\end{eqnarray*}
This post has been edited 1 time. Last edited by Tafi_ak, Apr 5, 2022, 8:21 AM
Reason: typo
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math31415926535
5617 posts
#20
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By CRT, we can break this into $\pmod{2}$ and $\pmod{5}.$ For $\pmod{2},$ note that $$|a_1-b_1| \equiv a_1+b_1 \pmod{2},$$since absolute value doesn't matter in $\pmod{2}.$ So our sum is just $1+2+...+1998 \equiv 1 \pmod{2}.$ For $\pmod{5},$ just notice that its always $1\pmod{5}$ for all of $|a_1-b_1|, |a_2-b_2|, ...$ so our answer is just $4\pmod{5}.$ Using CRT, we get an answer of $9\pmod{10}.$ $\blacksquare$
This post has been edited 1 time. Last edited by math31415926535, Dec 17, 2021, 3:41 AM
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th1nq3r
146 posts
#21
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Note that $|a - b| \equiv a - b \equiv a + b \pmod 2$ for all integers $a, b$.

The original sum then becomes $$|a_1 - b_1| + \cdots + |a_{999} - b_{999}| \equiv a_1 + b_1 + \cdots + a_{999} + b_{999} = 1 + \cdots + 1998 \equiv 1 \pmod 2. $$
Now we have that the sum $$|a_1 - b_1| + \cdots + |a_{999} - b_{999}| \equiv \underbrace{1 + \cdots + 1}_{\text{999 times}} = 999 \equiv 4 \pmod 5. $$
Hence this sum is $1 \pmod 2$ and $4 \pmod 5$, so by CRT we have that the sum is $9 \pmod {10}$, as desired. $\blacksquare$
This post has been edited 4 times. Last edited by th1nq3r, Aug 13, 2022, 5:03 AM
Reason: So many edits from the phone
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Alcumusgrinder07
95 posts
#22
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We will consider the sum mod 2 and mod 5 we see that since $6\equiv 1\pmod 2$ the sum of 999 terms has a remainder of 1 when divided by 2 similarly we see that $6\equiv1\pmod 5$ so the sum of 999 terms has a remainder of 4 when divided by 5 so by CRT we see that the sum has a remainder of 9 when divided by 10 which implies that the units digit is 9
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minusonetwelth
225 posts
#23
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Let the sum be $S$. Note that $S\equiv 999\cdot 1\mod 5\Longleftrightarrow S\equiv 4\mod 5$. Also note that $S$ has the same parity as $a_1+b_1+a_2+b_2+\ldots+a_{999}+b_{999}=1+\ldots+1998$ which clearly odd, so $S\equiv 1\mod 2$. Hence the remainder of $S\mod 10$ is unique by the Chinese Remainder Theorem, and it is easy to check that it is $9$ indeed.
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minusonetwelth
225 posts
#24
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Alcumusgrinder07 wrote:
We will consider the sum mod 2 and mod 5 we see that since $6\equiv 1\pmod 2$ the sum of 999 terms has a remainder of 1 when divided by 2 similarly we see that $6\equiv1\pmod 5$ so the sum of 999 terms has a remainder of 4 when divided by 5 so by CRT we see that the sum has a remainder of 9 when divided by 10 which implies that the units digit is 9

$6\equiv 1\mod 2$???
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Taco12
1757 posts
#25
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Note that the expression is $4 \pmod 5$, so it suffices to show that it is odd. But notice that we have $$\left|a_i-b_i\right| \equiv a_i-b_i \pmod 2,$$and we have $1+2+3+\dots+1998 \equiv 1 \pmod 2$, which finishes.
This post has been edited 1 time. Last edited by Taco12, Jan 15, 2023, 4:23 PM
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RedFireTruck
4223 posts
#26
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Since $1$ and $6$ differ by $5$, the ending digit is $9$ whenever there is an odd number of $i$ such that $|a_i-b_i|=1$ and is $4$ otherwise. There are $999$ odds and $999$ evens so there can't be an even number of $i$ such that $|a_i-b_i|=1$, because that would imply an even number of odds and an even number of evens. Therefore, the ending digit must be $9$.
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egg185
103 posts
#27
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MithsApprentice wrote:
Suppose that the set $\{1,2,\cdots, 1998\}$ has been partitioned into disjoint pairs $\{a_i,b_i\}$ ($1\leq i\leq 999$) so that for all $i$, $|a_i-b_i|$ equals $1$ or $6$. Prove that the sum \[ |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}|  \]ends in the digit $9$.

my teach gimme that as homework
Click to reveal hidden text
This post has been edited 3 times. Last edited by egg185, Aug 4, 2023, 6:37 AM
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shendrew7
796 posts
#28
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Suppose there are $x$ pairs with difference $1$. Then our sum is \[(999-x)+6x = 999+5x,\]which has units digit either $4$ or $9$.

Note that this sum also has the same parity as $1 + 2 + \ldots + 1998$, which is odd, since \[\lvert a_i - b_i \rvert \equiv a_i - b_i \equiv a_i + b_i \mod 2.\]
Thus our sum has units digit $9$. $\blacksquare$
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gracemoon124
872 posts
#29 • 1 Y
Y by salemath
Clearly, the desired sum is $4\pmod 5$, so it remains to show that it’s $1\pmod 2$.

Because $|a-b|\equiv a-b\equiv a+b\pmod 2$ for any integers $a$ and $b$, this means
\[|a_1-b_1|+|a_2-b_2|+\dots+|a_{999}-b_{999}|\equiv 1+2+\dots + 1998\equiv 1\pmod 2\]and we are done. $\square$
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mathmax12
6051 posts
#30
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We note that this sum is $\equiv 4\pmod{5},$ we need to prove that this is $1 \pmod{2},$ note that $|a-b|,$ has the same parity as $a+b,$ so we need to prove $a+b \equiv 1\pmod{2},$ so $|a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}|  \equiv 1+2+3...+1998 \equiv 1\pmod{2},$ hence proved $\blacksquare$
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joshualiu315
2534 posts
#31
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WLOG assume $a_i>b_i$ for each $i$.

Notice that

\[\sum a_i + \sum b_i = 999 \cdot 1999,\]
which is odd. Thus, in order to have integer values, we must have that $\sum a_i - \sum b_i$ is odd. Thus, the number of $6$'s, denote as $2k$, is even. We have

\[\sum a_i - \sum b_1 = 6 \cdot 2k + (999-2k) = 10k+999 \equiv 9 \pmod{10}. \ \square\]
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kamatadu
480 posts
#32 • 1 Y
Y by GeoKing
WLOG $a_i>b_i$ for all $i$.

Now note that,
\[ \displaystyle\sum |a_i-b_i| = \displaystyle\sum a_i - b_i \equiv \displaystyle\sum a_i + b_i = \dfrac{1998\cdot 1999}{2} \equiv 1 \pmod{2}. \]
Also,
\[ \displaystyle\sum a_i - b_i \equiv \displaystyle\sum 1 = 999 \equiv 4 \pmod{5}. \]
Thus combining these by CRT, we get that $\displaystyle\sum |a_i-b_i| \equiv 9 \pmod{10}$. :yoda:
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SouradipClash_03
166 posts
#33 • 1 Y
Y by GeoKing
Cute.
Solution
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LogiBobb
8 posts
#35 • 1 Y
Y by OronSH
$|a_i - b_i| = 1$ or $6$. Say $|a_i - b_i| = 1$ for $k$ of these $999$ pairs. The sum would then be equal to $k + 6(999 - k) = 5994 - 5k$. Note that a pair with difference $6$ has same parity and a pair with difference $1$ does not. All pairs of difference $6$ would give us an even number of even numbers and an even number of odd numbers. If $k$ was even, that suggests there are even numbers of odd numbers and even numbers of even numbers. However, since there is a total of $999$ even numbers and $999$ odd numbers, this cannot be true. So thus, $k$ is odd, and the sum would end with digit $9$.
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bjump
1026 posts
#36
Y by
solution
This post has been edited 1 time. Last edited by bjump, Jan 7, 2024, 4:00 PM
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think4l
344 posts
#37
Y by
Overkill solution, but proud of it :play_ball:

For the sake of simplicity, we will write all pairs $\{a_i, b_i\}$ such that $a_i < b_i$.

Let there be $x$ six-pairs $(a_i, b_i)$ such that $|a_i - b_i|=6$. Then, there are $999-x$ one-pairs such that $|a_i - b_i| = 1$. The given expression evaluates to $$6x + (999-x) = 5x + 999$$For this to be $9 \pmod{10}$, it suffices to show that $x$ is even.

Note that along with any six-pair $(n, n+6)$, in order to make successful pairings, we must have either exactly one of the six-pairs $$(n+1, n+7), (n+3, n+9), (n+5, n+11),$$or one of the following sets of three six-pairs $$\{(n+1, n+7), (n+2, n+8), (n+3, n+9)\}$$$$\{(n+1, n+7), (n+2, n+8), (n+5, n+11)\}$$$$\{(n+1, n+7), (n+4, n+10), (n+5, n+11)\}$$
Each option generates an even number of six-pairs, so $x$ is even, as desired $\blacksquare$
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AshAuktober
1006 posts
#38
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Modulo 2, $$ |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}| \equiv a_1 + \cdots + a_{999} + b_1 + \cdots + b_{999}$$$$= \frac{1998 \cdot 1999}{2} \equiv 1 \pmod{2}.$$Note that $|a_i - b_i| \equiv 1 \pmod{5} \forall i$, so $$ |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}| \equiv 1 + \cdots + 1 \equiv 999 \equiv 4 \pmod{5}.$$Now from the Chinese Remainder Theorem, $ |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}| \equiv 9 \pmod{10}$. $\square$
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gladIasked
648 posts
#39
Y by
We see that each $|a_i-b_i|\equiv 1\pmod 5$, and $999\equiv 4\pmod 5$, so $|a_1-b_1|+|a_2-b_2|+\dotsb +|a_{999}-b_{999}|\equiv 4\pmod 5$. Note that the $|a_i-b_i|\equiv a_i+b_i\pmod 2$, so $|a_1-b_1|+|a_2-b_2|+\dotsb +|a_{999}-b_{999}|\equiv \frac{1998\cdot 1999}{2}\equiv 1\pmod 2$. Thus, $|a_1-b_1|+|a_2-b_2|+\dotsb +|a_{999}-b_{999}|\equiv 9\pmod {10}$. $\blacksquare$
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megarnie
5606 posts
#40
Y by
Since each $|a_i - b_i|$ is $1\pmod 5$, the sum is clearly $4\pmod 5$. It suffices to show that the sum is odd. Clearly the absolute value when taking the parity, so the sum has the same parity as $\sum a_i - \sum b_i$. Let $x = \sum a_i$. We have $\sum (a_i + b_i) = \sum a_i + \sum b_i =  1 + 2 + \cdots + 1998$ is odd, so $\sum b_i$ has the opposite parity of $\sum a_i$, implying that $\sum a_i - \sum b_i$ is odd, as desired.
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megahertz13
3183 posts
#41
Y by
It is clear that $$|a_1-b_1|+|a_2-b_2|+\dotsb +|a_{999}-b_{999}|\equiv 4\pmod 5.$$The only thing left is to prove that $$|a_1-b_1|+|a_2-b_2|+\dotsb +|a_{999}-b_{999}|\equiv 1\pmod 2.$$However, $$|a_1-b_1|+|a_2-b_2|+\dotsb +|a_{999}-b_{999}|$$$$\equiv a_1-b_1+a_2-b_2+\dots+a_{999}-b_{999}$$$$\equiv a_1+b_1+a_2+b_2+\dots+a_{999}+b_{999}$$$$=\frac{1998\cdot 1999}{2}$$$$\equiv 1\pmod 2.$$
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Maximilian113
575 posts
#42
Y by
Let there be $x$ pairs that sum to $6,$ and $y$ pairs that sum to $1.$ Then $x+y=999 \implies 6x+y \equiv 4 \pmod 5.$ Meanwhile, if WLOG $a_i > b_i$ for all $i,$ then our sum equals $$1998 \cdot 1999/2 - 2\sum b_i \equiv 1 \pmod 2,$$so by CRT it ends in the digit $9,$ so we are done. QED
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quantam13
113 posts
#43
Y by
Call the sum $X$. Since the sum of all the numbers from 1 to 1998 is odd, $X$ is odd and hence it suffices to prove the $X\equiv 4$ modulo 5. But that is trivial since there are 999 terms in the sum, each of which are 1 modulo 5.
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NerdyNashville
14 posts
#44
Y by
Solution
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endless_abyss
47 posts
#45
Y by
Nice!

Claim: $\sum | a_i - b_i | \equiv 1$ (mod $2$)
$\sum | a_i - b_i | \equiv \sum a_i + b_i \equiv 1$ (mod $2$)
so the claim is proven.

Claim: $\sum | a_i - b_i | \equiv -1$ (mod $5$)
If there are x number of $|a_i - b_i| = 6 \equiv 1$ (mod $5$)
there are $999 - x$ $|a_i - b_i$
such that
$Ia_i - b_i| = 1 \equiv 1$ (mod $5$)
we're done after summing them up.

combining the two claims gives us $\sum | a_i - b_i | \equiv -1$ (mod $10$) as desired.

$\square$
:starwars:
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