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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Interesting inequality
sqing   0
a minute ago
Source: Own
Let $a,b\geq 0, 2a+2b+ab=5.$ Prove that
$$a+b^3+a^3b+\frac{101}{8}ab\leq\frac{125}{8}$$
0 replies
1 viewing
sqing
a minute ago
0 replies
Inequality em981
oldbeginner   19
N 6 minutes ago by sqing
Source: Own
Let $a, b, c>0, a+b+c=3$. Prove that
\[\sqrt{a+\frac{9}{b+2c}}+\sqrt{b+\frac{9}{c+2a}}+\sqrt{c+\frac{9}{a+2b}}+\frac{2(ab+bc+ca)}{9}\ge\frac{20}{3}\]
19 replies
oldbeginner
Sep 22, 2016
sqing
6 minutes ago
Inspired by RMO 2006
sqing   4
N 9 minutes ago by sqing
Source: Own
Let $ a,b >0  . $ Prove that
$$  \frac {a^{2}+1}{b+k}+\frac { b^{2}+1}{ka+1}+\frac {2}{a+kb}  \geq \frac {6}{k+1}  $$Where $k\geq 0.03 $
$$  \frac {a^{2}+1}{b+1}+\frac { b^{2}+1}{a+1}+\frac {2}{a+b}  \geq 3  $$
4 replies
sqing
Saturday at 3:24 PM
sqing
9 minutes ago
Inspired by 2025 Beijing
sqing   11
N 18 minutes ago by sqing
Source: Own
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Prove that
$$ab^2c^2d \leq 8$$$$a^2bcd^2 \leq 16$$$$ ab^3c^3d \leq \frac{2187}{128}$$$$ a^3bcd^3 \leq \frac{2187}{32}$$
11 replies
sqing
Saturday at 4:56 PM
sqing
18 minutes ago
nice ecuation
MihaiT   1
N Yesterday at 7:24 PM by Hello_Kitty
Find real values $m$ , s.t. ecuation: $x+1=me^{|x-1|}$ have 2 real solutions .
1 reply
MihaiT
Yesterday at 2:03 PM
Hello_Kitty
Yesterday at 7:24 PM
Linear algebra problem
Feynmann123   1
N Yesterday at 3:51 PM by Etkan
Let A \in \mathbb{R}^{n \times n} be a matrix such that A^2 = A and A \neq I and A \neq 0.

Problem:
a) Show that the only possible eigenvalues of A are 0 and 1.
b) What kind of matrix is A? (Hint: Think projection.)
c) Give a 2×2 example of such a matrix.
1 reply
Feynmann123
Yesterday at 9:33 AM
Etkan
Yesterday at 3:51 PM
Linear algebra
Feynmann123   6
N Yesterday at 1:09 PM by OGMATH
Hi everyone,

I was wondering whether when I tried to compute e^(2x2 matrix) and got the expansions of sinx and cosx with the method of discounting the constant junk whether it plays any significance. I am a UK student and none of this is in my School syllabus so I was just wondering…


6 replies
Feynmann123
Saturday at 6:44 PM
OGMATH
Yesterday at 1:09 PM
Local extrema of a function
MrBridges   2
N Yesterday at 11:36 AM by Mathzeus1024
Calculate the local extrema of the function $f:\mathbb{R}^2 \rightarrow \mathbb{R}$, $(x,y)\mapsto x^4+x^5+y^6$. Are they isolated?
2 replies
MrBridges
Jun 28, 2020
Mathzeus1024
Yesterday at 11:36 AM
Integral
Martin.s   3
N Yesterday at 10:52 AM by Figaro
$$\int_0^{\pi/6}\arcsin\Bigl(\sqrt{\cos(3\psi)\cos\psi}\Bigr)\,d\psi.$$
3 replies
Martin.s
May 14, 2025
Figaro
Yesterday at 10:52 AM
Reducing the exponents for good
RobertRogo   1
N Yesterday at 9:29 AM by RobertRogo
Source: The national Algebra contest (Romania), 2025, Problem 3/Abstract Algebra (a bit generalized)
Let $A$ be a ring with unity such that for every $x \in A$ there exist $t_x, n_x \in \mathbb{N}^*$ such that $x^{t_x+n_x}=x^{n_x}$. Prove that
a) If $t_x \cdot 1 \in U(A), \forall x \in A$ then $x^{t_x+1}=x, \forall x \in A$
b) If there is an $x \in A$ such that $t_x \cdot 1 \notin U(A)$ then the result from a) may no longer hold.

Authors: Laurențiu Panaitopol, Dorel Miheț, Mihai Opincariu, me, Filip Munteanu
1 reply
RobertRogo
May 20, 2025
RobertRogo
Yesterday at 9:29 AM
Sequence divisible by infinite primes - Brazil Undergrad MO
rodamaral   5
N Yesterday at 8:01 AM by cursed_tangent1434
Source: Brazil Undergrad MO 2017 - Problem 2
Let $a$ and $b$ be fixed positive integers. Show that the set of primes that divide at least one of the terms of the sequence $a_n = a \cdot 2017^n + b \cdot 2016^n$ is infinite.
5 replies
rodamaral
Nov 1, 2017
cursed_tangent1434
Yesterday at 8:01 AM
Reduction coefficient
zolfmark   2
N Yesterday at 7:42 AM by wh0nix

find Reduction coefficient of x^10

in(1+x-x^2)^9
2 replies
zolfmark
Jul 17, 2016
wh0nix
Yesterday at 7:42 AM
a^2=3a+2imatrix 2*2
zolfmark   4
N Yesterday at 2:44 AM by RenheMiResembleRice
A
matrix 2*2

A^2=3A+2i
A^3=mA+Li


i means identity matrix,

find constant m ، L
4 replies
zolfmark
Feb 23, 2019
RenheMiResembleRice
Yesterday at 2:44 AM
Find solution of IVP
neerajbhauryal   3
N Yesterday at 12:47 AM by MathIQ.
Show that the initial value problem \[y''+by'+cy=g(t)\] with $y(t_o)=0=y'(t_o)$, where $b,c$ are constants has the form \[y(t)=\int^{t}_{t_0}K(t-s)g(s)ds\,\]

What I did
3 replies
neerajbhauryal
Sep 23, 2014
MathIQ.
Yesterday at 12:47 AM
Prove that expression is always even.
shivangjindal   20
N Apr 18, 2025 by EVKV
Source: INMO 2014- Problem 2
Let $n$ be a natural number. Prove that,
\[ \left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor + \left\lfloor \sqrt{n} \right\rfloor \]
is even.
20 replies
shivangjindal
Feb 2, 2014
EVKV
Apr 18, 2025
Prove that expression is always even.
G H J
Source: INMO 2014- Problem 2
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shivangjindal
676 posts
#1 • 2 Y
Y by Adventure10, cubres
Let $n$ be a natural number. Prove that,
\[ \left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor + \left\lfloor \sqrt{n} \right\rfloor \]
is even.
This post has been edited 1 time. Last edited by ahaanomegas, Feb 2, 2014, 2:31 PM
Reason: Fixed LaTeX
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Altricono
42 posts
#2 • 5 Y
Y by Adventure10, Mango247, cubres, Japnoor2411, and 1 other user
Just induct. The idea is common.
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Rust
5049 posts
#3 • 7 Y
Y by makar, ashrith9sagar_1, Adventure10, Mango247, cubres, aidan0626, Japnoor2411
$\sum_{k=1}^n [\frac{n}{k}]=\sum_{k=1}^n\tau(k)$. $\tau(k)$ - is number of divisors of k.
If $k\not =m^2$ divisors $d,\frac{k}{d}$ different and $\tau(k)$ is even.
$\tau(k)$ is odd only if $k=m^2$. Therefore
$\sum_{k=1}^n[\frac{n}{k}]+[\sqrt n]=2[\sqrt n]\mod 2 =0\mod 2$.
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mathuz
1525 posts
#4 • 6 Y
Y by DanDumitrescu, Adventure10, Mango247, Math_DM, cubres, Japnoor2411
[Moderator: Please do not double-post and/or quote the post immediately before yours.]

$ \sum_{k=1}^n \left \lfloor \frac{ n }{ k }\right \rfloor=\sum_{k=1}^n \tau(k) $.
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Shravu
131 posts
#5 • 6 Y
Y by Pluto1708, ashrith9sagar_1, Adventure10, Mango247, cubres, Japnoor2411
Yes, Induction....

gif(n+1/k) > gif(n/k) if and only if k|n+1 gif(n+1/k) = gif(n/k)+1 so there is pairing

and if n+1 is a square gif(root(n+1)) > gif(root(n))
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Askr
181 posts
#6 • 8 Y
Y by Unknown-6174, Uagu, SHREYAS333, stevoo, Adventure10, Mango247, cubres, Japnoor2411
We prove the statement by Induction.

For n = 1, [1/1] + [1^(1/2)] = 1+1 = 2 (Base Case)
Lets assume that for n = n, [n/1] + [n/2] + [n/3] + .............. + [n/n] + [n^(1/2)] = even
Now, for n=n+1, we take two cases :

(i) n+1 is not a perfect square :
In this case n+1 has k factors (excluding 1 and n+1) where k is even.
If b is not a factor of n+1, then [(n+1)/b] = [n/b].
If b is a factor of n+1, then [(n+1)/b] = [n/b] +1. There will be k such cases.
Also, [(n+1)^(1/2)] = [n^(1/2)] since n+1 is not a perfect square.
therefore, [(n+1)/1] + [(n+1)/2] + [(n+1)/3] + ............... + [(n+1)/n] + [(n+1)/(n+1)] + [(n+1)^(1/2)]
= [n/1] + [n/2] + [n/3] + ............. + [n/n] + [n^(1/2)] + k + 2
= even (since k is even)

(ii) n+1 is a perfect square :
In this case n+1 has k factors (excluding 1 and n+1) where k is odd.
If b is not a factor of n+1, then [(n+1)/b] = [n/b].
If b is a factor of n+1, then [(n+1)/b] = [n/b] +1. There will be k such cases.
Also, [(n+1)^(1/2)] = [n^(1/2)] + 1 since n+1 is a perfect square.
therefore, [(n+1)/1] + [(n+1)/2] + [(n+1)/3] + ............... + [(n+1)/n] + [(n+1)/(n+1)] + [(n+1)^(1/2)]
= [n/1] + [n/2] + [n/3] + ............. + [n/n] + [n^(1/2)] + k + 3
= even (since k is odd)

We have proved both the cases.
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makar
1581 posts
#7 • 11 Y
Y by isomorphicdevikram, mathbuzz, Pluto1708, AlastorMoody, ashrith9sagar_1, Supercali, Adventure10, Mango247, cubres, TestX01, Japnoor2411
shivangjindal wrote:
Let $n$ be a natural number. Prove that,
\[ \left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor + \left\lfloor \sqrt{n} \right\rfloor \]
is even.

By counting number of Lattice point in first quadrant and below or on the curve $xy=n$ in two different ways, we get:

$\left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor =2\sum _{k=1}^{\left\lfloor \sqrt{n} \right\rfloor }\left\lfloor \frac{n}{k} \right\rfloor -\left\lfloor \sqrt{n} \right\rfloor^2 $ :!:

Now

$\left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor + \left\lfloor \sqrt{n} \right\rfloor=\underbrace{2\sum _{k=1}^{\left\lfloor \sqrt{n} \right\rfloor }\left\lfloor \frac{n}{k} \right\rfloor}_\text{Even} -\underbrace{ \left\lfloor \sqrt{n} \right\rfloor\left(\left\lfloor \sqrt{n} \right\rfloor-1\right)}_\text{Even}=\text{Even} $ :)
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mathbuzz
803 posts
#8 • 8 Y
Y by AHKR, div5252, maths_arka, Adventure10, Mango247, Math_DM, cubres, Japnoor2411
Non-inductive solution --
Lemma -- $[\frac{n}{1}]+......+[\frac{n}{n}]=d(1)+d(2)+....d(n)$
Proof -- the number of elements in S={1,2,...,n} ,which are divisible by $q$ is $[\frac{n}{q}]$.
Now , the rest is trivial
end of lemma
Back to the main problem --
Now , note that ,
$\sum[\frac{n}{k}]+[\sqrt{n}] =d(1)+d(2)+.....+d(n)+[\sqrt{n}]=f(n)$ (say)
Now , note that , $d(k)=1(mod2)$ iff $k$ is a perfect square ..
There are only $[\sqrt{n}]$ perfect squares in {1,2,...,n}.
hence , $f(n)=[\sqrt{n}]+[\sqrt{n}]=0(mod2)$. :D
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BSJL
641 posts
#9 • 4 Y
Y by Adventure10, Mango247, cubres, Japnoor2411
We just need to notice that the following two points:

1. For $ n=1 $, It is true.

2.$ ([\frac{n+1}{1}]+......+[\frac{n+1}{n+1}]+[\sqrt{n+1}])-([\frac{n}{1}]+......+[\frac{n}{n}]+[\sqrt{n}])=2d(n+1) $
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AlastorMoody
2125 posts
#12 • 4 Y
Y by Arhaan, Adventure10, cubres, Japnoor2411
INMO 2014 P2 wrote:
Let $n$ be a natural number. Prove that,
\[ \left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor + \left\lfloor \sqrt{n} \right\rfloor \]is even.
Solution: Notice, for $n,k \in \mathbb{N}$, $\left \lfloor \frac{n}{k} \right \rfloor$ denotes the number of positive multiples of $k$ below $\left \lfloor \frac{n}{k} \right \rfloor$. Also, any multiple in the form of $k \times \ell \le n$ will be counted twice, once in $\left \lfloor \frac{n}{k} \right \rfloor$ and in $ \left \lfloor \frac{n}{\ell} \right \rfloor$ where $k, \ell$ are distinct. Only Numbers that can be represented in the form of squares (eg: $n \times n$) are not counted twice, but adding a $\lfloor {\sqrt{n}} \rfloor$ to the expression makes all the counts twice and hence the resultant is even $\qquad \blacksquare$
This post has been edited 1 time. Last edited by AlastorMoody, Sep 5, 2019, 4:52 PM
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Mogmog8
1080 posts
#13 • 3 Y
Y by centslordm, cubres, Japnoor2411
It is well known that $\sum_{i=1}^n\left\lfloor\frac{n}{i}\right\rfloor=\sum_{i=1}^n d(i),$ where $d(i)$ denotes the number of factors of $i.$ Notice $d(i)\equiv 1\pmod{2}$ if and only if $i$ is a perfect square. We know there are $\left\lfloor\sqrt{n}\right\rfloor$ perfect squares from $1$ to $n$ so $$\left(\sum_{i=1}^n d(i)\right)+\left\lfloor\sqrt{n}\right\rfloor\equiv 2\left\lfloor\sqrt{n}\right\rfloor\equiv 0\pmod{2}.$$$\square$
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minusonetwelth
225 posts
#14 • 3 Y
Y by panche, cubres, Japnoor2411
We will first prove the following well-known result:
Claim:
If $d(n)$ is the number of divisors of a positive integer $n$, then
$$d(1)+d(2)+\ldots+d(n)=\lfloor\frac{n}{1}\rfloor+\lfloor\frac{n}{2}\rfloor+\ldots+\lfloor\frac{n}{n}\rfloor$$Proof of the claim:
\begin{align*}
			d(1)+d(2)+\ldots+d(n)&=\sum_{k=1}^{n}{\sum_{i|k}{1}}\\
			&=\sum_{i=1}^{n}{\sum_{\substack{1\le j\le n\\ i|j}}{1}}\\
			&=\sum_{i=1}^{n}{\lfloor\frac{n}{i}\rfloor}=\lfloor\frac{n}{1}\rfloor+\lfloor\frac{n}{2}\rfloor+\ldots+\lfloor\frac{n}{n}\rfloor
		\end{align*}as desired.

Now we proceed with induction. The base case $n=1$ is clear. Assume that $2| d(1)+\ldots+d(k)+\lfloor\sqrt{k}\rfloor$. Then for $k+1$
\begin{align*}
		2&|d(1)+d(2)+\ldots+d(k)+d(k+1)+\lfloor\sqrt{k+1}\rfloor\Longleftrightarrow\\
		2&|d(1)+d(2)+\ldots+d(k)+d(k+1)+\lfloor\sqrt{k+1}\rfloor+\lfloor\sqrt{k}\rfloor-\lfloor\sqrt{k}\rfloor\Longleftrightarrow\\
		2&| d(k+1)+\lfloor\sqrt{k+1}\rfloor-\lfloor\sqrt{k}\rfloor
	\end{align*}If $k+1$ is a square, then $d(k+1)$ is odd, while $\lfloor\sqrt{k+1}\rfloor-\lfloor\sqrt{k}\rfloor=1$. If $k+1$ is not a square, $d(k+1)$ is even, and $\lfloor\sqrt{k+1}\rfloor-\lfloor\sqrt{k}\rfloor=0$, finishing the proof.
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lifeismathematics
1188 posts
#15 • 2 Y
Y by Math_DM, Japnoor2411
first of all we have $\left \lfloor \frac{n}{1} \right \rfloor +\left \lfloor \frac{n}{2}  \right \rfloor +\cdots+\left \lfloor \frac{n}{n}  \right \rfloor=\tau(1)+\tau(2)+\cdots+\tau(n)$

clearly, $\tau(j) \equiv 0 \pmod 2$ if $ j$ is not a perfect square, and $\equiv 1 \pmod 2$ if it is a perfect square. We notice in set $\{1,2 , \cdots , n\}$ we have $\lfloor \sqrt{n} \rfloor$ many perfect squares, so we have \[ \left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor + \left\lfloor \sqrt{n} \right\rfloor \]$\equiv 2 \lfloor \sqrt{n} \rfloor \equiv 0 \pmod 2$ as there are just $\lfloor \sqrt{n} \rfloor$ many perfect squares in $[n]$ so done. $\blacksquare$
This post has been edited 1 time. Last edited by lifeismathematics, Aug 11, 2023, 12:52 PM
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AshAuktober
1009 posts
#16 • 1 Y
Y by Japnoor2411
Basic idea:
d(1) + d(2) + \cdots + d(n) + [\sqrt{n}
and parity changes only when n is a square
where parity of [sqrt(n)] changes too
so parity of expression is invariant
now we see at n = 1 its even
so its always even
qed
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Siddharthmaybe
117 posts
#17 • 1 Y
Y by Japnoor2411
[n/1] + [n/2] +.... + [n/n] = d(1) + d(2) +.... d(n) (lemma)
no. of divisors of a square is odd
Case 1->
there are an even number of naturals <= [rootn] (the squares till n are even in no. )
<=> [rootn] is even and the sum of all d(i) from 1 to [rootn] is obv even and we are done
similar logic for the other case works :)
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cursed_tangent1434
642 posts
#18 • 1 Y
Y by Japnoor2411
This problem really made me appreciate how theory dependent Olympiad number theory really was. Knowing that one trick really made solving the problem much easier. The following lemma is essential.

Lemma : For all positive integers $n$ and $k$,
\[\left\lfloor \frac{n}{k} \right\rfloor - \left\lfloor \frac{n-1}{k} \right\rfloor = \begin{cases}1 & k \mid n \\ 0 & k \nmid n\end{cases}\]


Proof : The proof is clear. $[\left\lfloor \frac{n}{k} \right\rfloor  > \left\lfloor \frac{n-1}{k} \right\rfloor$ if and only if we jump to a new multiple of $k$.

Now, we proceed via induction. Note how the problem is clear for $n=1$. Now, assuming the problem when $n=k$ we show the result when $n=k+1$. Note,
\begin{align*}
\left(\sum_{i=1}^{k+1} \left \lfloor \frac{k+1}{i} \right \rfloor + \sqrt{k+1} \right) - \left(\sum_{i=1}^{k} \left \lfloor \frac{k}{i} \right \rfloor + \sqrt{k} \right ) &= d(k+1) + \sqrt{k+1}- \sqrt{k}
\end{align*}
Now, if $k+1$ is not a perfect square, $d(k+1)$ is even, and $\sqrt{k+1}=\sqrt{k}$. Thus, the result holds when $n=k+1$ as well. Further if $k+1$ is a perfect square, $\sqrt{k+1}-\sqrt{k}=1$ but $d(k+1)$ is odd, and thus again the result holds for $n=k+1$ and we are done.
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eg4334
636 posts
#19 • 1 Y
Y by Japnoor2411
It rewrites into $d(1)+d(2) + \dots + d(n) + \lfloor \sqrt{n} \rfloor$. All the $d(i)$ are even except the squares but the latter term makes up for it by adding one for each square.
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LeYohan
56 posts
#21 • 1 Y
Y by Japnoor2411
It's known that $\left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor = d(1) + d(2) + \cdots + d(n)$ where $d(x)$ is the number of divisors of $x$, so the problem is now equivalent to showing that $d(1) + d(2) + \cdots + d(n) + \left\lfloor \sqrt{n} \right\rfloor$ is even.

Let's express $n$ as $k^2 + l$ where $0 \leq l \leq 2k$, and let's consider two cases:

Case 1: $k$ is odd.

In this case, there is an odd number of squares bewteen $1$ and $n$ inclusive, and notice that it does not matter where $n$ lies between $k^2$ and $(k+1)^2$ since we remember that $d(x)$ is odd if and only if $x$ is a square. This means that $d(1) + d(2) + \cdots + d(n)$ is odd, and $\left\lfloor \sqrt{n} \right\rfloor = k$ which is odd, so the total sum is even. $\square$

Case 2: $k$ is even.

Similarly, we don't care where $n$ lies between $k^2$ and $(k+1)^2$ and notice that there is an even number of squares between $1$ and $n$ inclusive. Therefore, $d(1) + d(2) + \cdots + d(n)$ is even and $\left\lfloor \sqrt{n} \right\rfloor = k$, which is also even, so the total sum is even. $\square$
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jasperE3
11384 posts
#22 • 1 Y
Y by Japnoor2411
shivangjindal wrote:
Let $n$ be a natural number. Prove that,
\[ \left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor + \left\lfloor \sqrt{n} \right\rfloor \]is even.

Claim: $\tau(n+1)+\left\lfloor\sqrt{n+1}\right\rfloor-\left\lfloor\sqrt n\right\rfloor$ is even for all $n\ge0$

Let $\left\lfloor\sqrt{n+1}\right\rfloor=k$.

Case 1: $n+1$ is a square
We claim that $\left\lfloor\sqrt{n+1}\right\rfloor-\left\lfloor\sqrt n\right\rfloor=1$.

Write $n+1=k^2$. We have, since $k\ge0$:
$$k^2-1\ge(k-1)^2\Rightarrow n\ge(k-1)^2$$as well as $n<n+1=k^2$. Then $k-1\le\sqrt n<k$ so we must have $\left\lfloor\sqrt n\right\rfloor=k-1$, so:
$$\left\lfloor\sqrt{n+1}\right\rfloor-\left\lfloor\sqrt n\right\rfloor=1.$$
Then since $\tau(n+1)$ is odd, our claim is proven for this case.

Case 2: $n+1$ isn't a square
We claim that $\left\lfloor\sqrt{n+1}\right\rfloor-\left\lfloor\sqrt n\right\rfloor=0$.

By the definition of $k$, we have $k^2\le n+1<(k+1)^2$, but since $n+1$ isn't a square, this inequality is strict on both sides. Then $k^2\le n<n+1<(k+1)^2$ so $\left\lfloor\sqrt n\right\rfloor=k$, hence $\left\lfloor\sqrt{n+1}\right\rfloor-\left\lfloor\sqrt n\right\rfloor=0$.

Then since $\tau(n+1)$ is odd, our claim is proven.



We finish the problem by summing this from $0$ to $n-1$, giving that:
$$\sum_{i=1}^n\tau(i)+\left\lfloor\sqrt n\right\rfloor$$is even, and we're done since $\sum_{i=1}^n\tau(i)=\sum_{i=1}^n\left\lfloor\frac ni\right\rfloor$.
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alexanderhamilton124
400 posts
#23 • 3 Y
Y by Nobitasolvesproblems1979, Japnoor2411, L13832
Breh
Let us induct. We are clearly done for \( n = 1 \), so suppose we are done for \( n \in \{1, \dots, k\} \). Let \( f(n) \) be our expression. Note that

\[
\left\lfloor \frac{k + 1}{x} \right\rfloor - \left\lfloor \frac{k}{x} \right\rfloor = 1 \text{ if } x \mid (k + 1), \text{ and } 0 \text{ otherwise}.
\]
Hence:

\[
f(k + 1) - f(k) = d(k + 1) + \left\lfloor \sqrt{k + 1} \right\rfloor - \left\lfloor \sqrt{k} \right\rfloor
\]
If \( k + 1 \) is not a perfect square, \( d(k + 1) \) is even and

\[
\left\lfloor \sqrt{k + 1} \right\rfloor = \left\lfloor \sqrt{k} \right\rfloor
\]
which gives \( f(k + 1) - f(k) \) even, so \( f(k + 1) \) even as well, and we're done.

If \( k + 1 \) is a perfect square, \( d(k + 1) \) is odd and

\[
\left\lfloor \sqrt{k + 1} \right\rfloor = \left\lfloor \sqrt{k} \right\rfloor + 1,
\]
so \( f(k + 1) - f(k) \) is even again and we're done.
This post has been edited 1 time. Last edited by alexanderhamilton124, Mar 17, 2025, 10:19 PM
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EVKV
71 posts
#24
Y by
Trivial nt in INMO what a suprise :joy:

Just induction and use the fact that d(n)= odd iff n is a perfect square
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