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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
Factorials divide
va2010   37
N 12 minutes ago by ND_
Source: 2015 ISL N2
Let $a$ and $b$ be positive integers such that $a! + b!$ divides $a!b!$. Prove that $3a \ge 2b + 2$.
37 replies
1 viewing
va2010
Jul 7, 2016
ND_
12 minutes ago
IMO Shortlist 2011, Number Theory 2
orl   24
N 15 minutes ago by ezpotd
Source: IMO Shortlist 2011, Number Theory 2
Consider a polynomial $P(x) =  \prod^9_{j=1}(x+d_j),$ where $d_1, d_2, \ldots d_9$ are nine distinct integers. Prove that there exists an integer $N,$ such that for all integers $x \geq N$ the number $P(x)$ is divisible by a prime number greater than 20.

Proposed by Luxembourg
24 replies
orl
Jul 11, 2012
ezpotd
15 minutes ago
Inequality in triangle
Nguyenhuyen_AG   3
N 23 minutes ago by Nguyenhuyen_AG
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that
\[\frac{1}{(a-4b)^2}+\frac{1}{(b-4c)^2}+\frac{1}{(c-4a)^2} \geqslant \frac{1}{ab+bc+ca}.\]
3 replies
Nguyenhuyen_AG
Today at 6:17 AM
Nguyenhuyen_AG
23 minutes ago
Problem 1
randomusername   73
N 35 minutes ago by ND_
Source: IMO 2015, Problem 1
We say that a finite set $\mathcal{S}$ of points in the plane is balanced if, for any two different points $A$ and $B$ in $\mathcal{S}$, there is a point $C$ in $\mathcal{S}$ such that $AC=BC$. We say that $\mathcal{S}$ is centre-free if for any three different points $A$, $B$ and $C$ in $\mathcal{S}$, there is no points $P$ in $\mathcal{S}$ such that $PA=PB=PC$.

(a) Show that for all integers $n\ge 3$, there exists a balanced set consisting of $n$ points.

(b) Determine all integers $n\ge 3$ for which there exists a balanced centre-free set consisting of $n$ points.

Proposed by Netherlands
73 replies
randomusername
Jul 10, 2015
ND_
35 minutes ago
x is rational implies y is rational
pohoatza   44
N an hour ago by ezpotd
Source: IMO Shortlist 2006, N2, VAIMO 2007, Problem 6
For $ x \in (0, 1)$ let $ y \in (0, 1)$ be the number whose $ n$-th digit after the decimal point is the $ 2^{n}$-th digit after the decimal point of $ x$. Show that if $ x$ is rational then so is $ y$.

Proposed by J.P. Grossman, Canada
44 replies
pohoatza
Jun 28, 2007
ezpotd
an hour ago
Multiplicative function
Tales   37
N an hour ago by ezpotd
Source: IMO Shortlist 2004, number theory problem 2
The function $f$ from the set $\mathbb{N}$ of positive integers into itself is defined by the equality \[f(n)=\sum_{k=1}^{n} \gcd(k,n),\qquad n\in \mathbb{N}.\]
a) Prove that $f(mn)=f(m)f(n)$ for every two relatively prime ${m,n\in\mathbb{N}}$.

b) Prove that for each $a\in\mathbb{N}$ the equation $f(x)=ax$ has a solution.

c) Find all ${a\in\mathbb{N}}$ such that the equation $f(x)=ax$ has a unique solution.
37 replies
Tales
Mar 23, 2005
ezpotd
an hour ago
NICE INEQUALITY
Kyleray   3
N an hour ago by sqing
Let's $a,b,c>0$. Prove:
$$(\frac{a}{b+c}+\frac{b}{c+a})(\frac{b}{c+a}+\frac{c}{a+b})(\frac{c}{a+b}+\frac{a}{b+c})\geq \frac{(a+b+c)^2}{3(ab+bc+ca)}$$$\text{P/S: No mapple, please :(}$
3 replies
Kyleray
Mar 11, 2021
sqing
an hour ago
Tough inequality
TUAN2k8   4
N an hour ago by cazanova19921
Source: Own
Let $n \ge 2$ be an even integer and let $x_1,x_2,...,x_n$ be real numbers satisfying $x_1^2+x_2^2+...+x_n^2=n$.
Prove that
$\sum_{1 \le i < j \le n} \frac{x_ix_j}{x_i^2+x_j^2+1} \ge \frac{-n}{6}$
4 replies
TUAN2k8
May 28, 2025
cazanova19921
an hour ago
Easy Diophantne
anantmudgal09   20
N 2 hours ago by Adywastaken
Source: India Practice TST 2017 D1 P2
Find all positive integers $p,q,r,s>1$ such that $$p!+q!+r!=2^s.$$
20 replies
anantmudgal09
Dec 9, 2017
Adywastaken
2 hours ago
Converse of a classic orthocenter problem
spartacle   43
N 2 hours ago by ihategeo_1969
Source: USA TSTST 2020 Problem 6
Let $A$, $B$, $C$, $D$ be four points such that no three are collinear and $D$ is not the orthocenter of $ABC$. Let $P$, $Q$, $R$ be the orthocenters of $\triangle BCD$, $\triangle CAD$, $\triangle ABD$, respectively. Suppose that the lines $AP$, $BQ$, $CR$ are pairwise distinct and are concurrent. Show that the four points $A$, $B$, $C$, $D$ lie on a circle.

Andrew Gu
43 replies
1 viewing
spartacle
Dec 14, 2020
ihategeo_1969
2 hours ago
Symmetric points part 2
CyclicISLscelesTrapezoid   22
N 2 hours ago by ihategeo_1969
Source: USA TSTST 2022/6
Let $O$ and $H$ be the circumcenter and orthocenter, respectively, of an acute scalene triangle $ABC$. The perpendicular bisector of $\overline{AH}$ intersects $\overline{AB}$ and $\overline{AC}$ at $X_A$ and $Y_A$ respectively. Let $K_A$ denote the intersection of the circumcircles of triangles $OX_AY_A$ and $BOC$ other than $O$.

Define $K_B$ and $K_C$ analogously by repeating this construction two more times. Prove that $K_A$, $K_B$, $K_C$, and $O$ are concyclic.

Hongzhou Lin
22 replies
CyclicISLscelesTrapezoid
Jun 27, 2022
ihategeo_1969
2 hours ago
Periodicity of factorials
Cats_on_a_computer   0
2 hours ago
Source: Thrill and challenge of pre-college mathematics
Let a_k denote the first non zero digit of the decimal representation of k!. Does the sequence a_1, a_2, a_3, … eventually become periodic?
0 replies
Cats_on_a_computer
2 hours ago
0 replies
Cyclic Quad. and Intersections
Thelink_20   11
N 2 hours ago by americancheeseburger4281
Source: My Problem
Let $ABCD$ be a quadrilateral inscribed in a circle $\Gamma$. Let $AC\cap BD=E$, $AB\cap CD=F$, $(AEF)\cap\Gamma=X$, $(BEF)\cap\Gamma=Y$, $(CEF)\cap\Gamma=Z$, $(DEF)\cap\Gamma=W$, $XZ\cap YW=M$, $XY\cap ZW=N$. Prove that $MN$ lies over $EF$.
11 replies
Thelink_20
Oct 29, 2024
americancheeseburger4281
2 hours ago
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   15
N 2 hours ago by math90
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
15 replies
OgnjenTesic
May 22, 2025
math90
2 hours ago
USAMO 2003 Problem 1
MithsApprentice   70
N May 27, 2025 by endless_abyss
Prove that for every positive integer $n$ there exists an $n$-digit number divisible by $5^n$ all of whose digits are odd.
70 replies
MithsApprentice
Sep 27, 2005
endless_abyss
May 27, 2025
USAMO 2003 Problem 1
G H J
G H BBookmark kLocked kLocked NReply
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AngeloChu
471 posts
#59 • 1 Y
Y by cubres
we will use induction
we first have that there exists $(2x_0+1)*10^0$ that is divisible by $5^1$
next, assume there exists $(2x_0+1)*10^{n-1}+...+(2x_n+1)*10^0$ divisible by $5^n$
multiply by $10$ to get $2(x_1*10^n+...+x_n*10)+\frac{10^{n+1}-10}{9}$
this is divisible by $5^{n+1}$
add one to get $2(x_1*10^n+...+x_n*10)+\frac{10^{n+1}-1}{9} \equiv 1 (\text{mod }5^{n+1})$
we need $2(y_1*10^n+...+y_n) \equiv -1 (\text{mod }5^{n+1})$
the existence of this is trivial since $5^{n+1}-1$ is even
therefore, we are done
This post has been edited 2 times. Last edited by AngeloChu, Jan 27, 2024, 2:54 AM
Reason: fixed latex fr this time
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Carolstar9
827 posts
#60 • 2 Y
Y by busy-beaver, cubres
Apologies for adding another solution to the long list already present here.

Look at the $n$-digit decimal representations of the numbers $S = \left\{k\cdot 5^n\,\mid\, 0\leq k < 2^n\right\}$ [i.e., if the number has less than $n$ digits, then add appropriate number of zeroes at the beginning]. Define a map $f : S \to \{0,1\}^n$ as follows:
$ f(\overline{a_1 a_2 \cdots a_n}) = (b_1, b_2, \cdots , b_n) $, where $ b_i = 0 $ if $ a_i $ is even and $ b_i = 1 $ if $ a_i $ is odd. If $ f(k\cdot 5^n) = f(l\cdot 5^n) $ where $ 0 \leq k \leq l < 2^n$ then $ (k-l) \cdot 5^n = \sum_{i=0}^{n-1} c_i \cdot 10^i $ for some $ c_i \in \{-8, -6, -4, -2, 0, 2, 4, 6, 8\} $. If any of the $ c_i $ are nonzero, let $ i $ be the smallest one. Then, the highest power of $5$ dividing $ \sum_{i=0}^{n-1} c_i \cdot 10^i $ is $i$, which is a contradiction. Therefore, $ k = l $, which implies that $f$ is injective. Since $|S| = 2^n = |\{ 0,1\}^n| $, we get that $f$ is surjective as well.
Looking at the pre image of $(1, 1,\cdots , 1)$, we are done.
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EpicBird08
1756 posts
#61 • 1 Y
Y by cubres
sus this was easier than i thought it was

We will construct such a number inductively. Let $a_n$ be our constructed number with $n$ digits divisible by $5^n$ with all odd digits. For the base case $n = 1,$ take $a_1 = 5.$ Now suppose we have our number $a_k$ that is divisible by $5^k.$ Then we will add a digit to the beginning of $a_k$ to make $a_{k+1}.$ To add the digit at the beginning, we need to add one of $10^k, 3 \cdot 10^k, 5 \cdot 10^k, 7 \cdot 10^k,$ or $9 \cdot 10^k.$ Adding any of these will not change whether or not the expression is divisible by $5^k.$ Moreover, if we let $u = 2^k,$ the possible numbers we can add become $u \cdot 5^k, 3u \cdot 5^k, 5u \cdot 5^k, 7u \cdot 5^k,$ and $9u \cdot 5^k.$ Note that $\gcd(u, 5) = 1,$ so taking mod $5^{k+1}$ gives the list $0, 5^k, 2 \cdot 5^k, 3 \cdot 5^k, 4 \cdot 5^k$ in some order. Then if $a_n \equiv z \cdot 5^k \pmod{5^{k+1}},$ we just add the digit which gives the residue $(5-z) \cdot 5^k \pmod{5^{k+1}},$ which will give $a_{k+1}.$ The induction is complete, so we have solved the problem.
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AshAuktober
1013 posts
#62 • 1 Y
Y by cubres
We prove this by induction.
Base case ($n = 1$) is trivial, consider 5.
Now say we have a number $k_{n-1} =\overline{ a_{n-1}\cdots a_1}$, all of whose digits are odd and which is a multiple of $5^{n-1}$, i. e. $k_{n-1} = p5^{n-1}$.
Now consider $a_n$ such that $5 \mid 2^{n-1}a_n + p$. Clearly there is a unique residue $\pmod{5}$ satisfying this, and so two such numbers in $\{0, 1, \cdots, 9\}$; one odd and one even. Pick the $a_n$ which is odd. Then the number $k_n = \overline{a_n \cdots a_1}$ works.$\square$
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lnzhonglp
120 posts
#63 • 1 Y
Y by cubres
Induct on $n$, base case $n=1$ is $5$. Now suppose there exists an $n$-digit number $a$ all of whose digits are odd. We will add $k10^n$, where $k \in \{1, 3, 5, 7, 9\}$. $1, 3, 5, 7, 9$ are distinct $\bmod \ 5$, therefore one will result in $a + k10^n \equiv 0 \pmod {5^{n+1}}$.
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v4n1lla
5 posts
#64 • 2 Y
Y by sabkx, cubres
This problem appeared in the 2024 Singapore Mathmatical Olympiad Open Round 2.
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PEKKA
1850 posts
#65 • 1 Y
Y by cubres
We will use induction.
Base case: $n=1.$ The number $5$ works.
Inductive step: Assume there exists a number $N$ with $k$ digits, each digit odd that is a multiple of $5^k.$
The possible $k+1$ digit numbers are $N+10^k,N+3 \left(10^k \right) \dots N+9 \left(10^k \right).$
Each of those numbers can be factored as
$5^k(R+2^k), 5^k(R+3\left(2^k \right) \dots 5^k(R+9\left(2^k\right)$.
Here, $R$ is the number $\frac{N}{5^k}.$
Each of the numbers of the form $R+2^k, R+3\cdot 2^k, \dots$ are equivalent to a different residue modulo 5, so one of them must be a multiple of $5$.
Therefore, there exists an odd number we can "place" in front of $N$ to create a $k+1$ digit number that is a multiple of $5^{k+1}$ with all digits odd.
The induction is complete and the desired statement holds.
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dudade
139 posts
#66 • 1 Y
Y by cubres
We will prove this with induction.

Base Case. $n = 1$.
We can consider the odd $1$-digit number $5$ which is divisible by $5^1$.

Induction. Suppose the conditions hold for $k = n$, we will show it works for $k = n + 1$.
Suppose $T \cdot 5^n$ is a $n$-digit number whose digits are all odd. Then, consider the following numbers:
\begin{align*}
1 \cdot 10^n + T \cdot 5^n, \quad 3 \cdot 10^n + T \cdot 5^n, \quad 5 \cdot 10^n + T \cdot 5^n, \quad 7 \cdot 10^n + T \cdot 5^n, \quad 9 \cdot 10^n + T \cdot 5^n.
\end{align*}Factoring out $5^n$ from these numbers yields
\begin{align*}
1 \cdot 2^n + T, \quad 3 \cdot 2^n + T, \quad 5 \cdot 2^n + T, \quad 7 \cdot 2^n + T, \quad 9 \cdot 2^n + T.
\end{align*}Suppose $p$ and $q$ are unique values selected from this list. Then, $p - q = r \cdot 2^n$ for some $r$. Notice $r$ cannot be divisible by $5$, thus $p$ and $q$ are unique values modulo $5$. Therefore, these five values are all unique modulo $5$. So, one of them must be divisible by $5$, as desired.
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surpidism.
10 posts
#67 • 1 Y
Y by cubres
We use induction on $n$.

Base case is obvious.

For $n= k$, assume there exits an integer $N$ such that $5^k \vert N$, and $N$ is $k$-digit number with all digits being odd.

Now, for $n = k+1$, consider the numbers $i \times 10^k + N$, $i = 1, 3, 5, 7, 9$.
Note that $5^k \vert i \times 10^k + N$, which can be written as $i \times 10^k + N = 5^k \left(  i \times 2^k + \frac{N}{5^k} \right)$.
Observe that the numbers $ i \times 2^k + \frac{N}{5^k}$ give five different remainders when divided by $5$, so one of them must be divisibly by $5$.
This implies $i \times 10^k + N$ is divisibly by $5^{k+1}$ for one $i \in \{1, 3, 5, 7, 9 \}$, and has $k+1$ odd digits, as required. $\square$
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Marcus_Zhang
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#68 • 1 Y
Y by cubres
5 min solve and write-up
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eg4334
636 posts
#69
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Proceed using induction with trivial base case. Assume for $n$ a valid integer $x$ exists. Now consider appending $x$ to $a = 1, 3, 5, 7, 9$. $x \equiv 0 \pmod{5^n}$, so let $x \equiv y \cdot 5^n \pmod{5^{n+1}}$ where $y = 0, 1, 2, 3, 4$. Now we need to solve $y+2^n a \equiv 0 \pmod{5}$ which obviously has a solution which finishes.
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L13832
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#70 • 1 Y
Y by alexanderhamilton124
When $n=1$, $5$ works and for $n=2$, $25$ works. Now that we are done with the base case, by induction hypothesis we assume that $N$ satisfies the property for $5^k$ then one of $N+i\cdot 10^k$, $i\in\{1,3,5,7,9\}$ is clearly divisible by $5^{k+1}$.
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Mamadi
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#71
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By taking a look at the examples for \( n = 1, 2, \dots, 5 \):
\[
n = 1: 5,
\]\[
n = 2: 75,
\]\[
n = 3: 375,
\]\[
n = 4: 9375,
\]\[
n = 5: 59375
\]we can assume that This problem could be solved by induction for the first digit from the left.

we can assume that for \( k \in \{1, 2, \dots, n-1\} \) we have a \( k \) digit number divisible by \( 5^k \)

for \( k = n-1 \), we call the \( n-1 \) digit number \( a \).

we know that \( a = 5^{n-1} \cdot c \)
we want to add \( b \cdot 10^{n-1} \) such that \( b \cdot 2^{n-1} + c \equiv 0 \pmod{5} \) and \( b \) is an odd number

we know that using odd numbers we can achieve any number (mod 5)

so lets assume \( 2^{n-1} \equiv d \pmod{5} \)

so we can add \( b \cdot 10^{n-1} \) to \( a \) such that \( bd \equiv -c \pmod{5} \)

now we have a new \( n \) digit number: \( b \cdot 10^{n-1} + a = b \cdot 2^{n-1} \cdot 5^{n-1} + 5^{n-1} \cdot c = 5^{n-1}(b \cdot 2^{n-1} + c) \) such that \( b \cdot 2^{n-1} + c \equiv 0 \pmod{5} \)

done.
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de-Kirschbaum
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#72
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For $n=1$ we can simply take the number to be $5$. Let us assume that there exists a positive integer with all odd digits divisible by $5^k$ for some $k$, then consider the case of $k+1$. Let the integer we had before be expressed as $5^kt$, then we claim that we can always add an odd number to the front of it to make it divisible by $5^{k+1}$. Indeed we want $10^ka+5^kt=5^k(2^ka+t) \equiv 0 \mod{5^{k+1}} \implies 2^ka+t \equiv 0 \mod{5} \implies a \equiv -3^kt \mod{5}$. If it is $0$ we take $a=5$, if it is $1,3$ we take $a=1,3$, if it is $2,4$ we take $a=7,9$. Thus we get a number of the desired qualities.
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endless_abyss
69 posts
#73
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Nice, induction insta-solves this problem.

Just start off by noticing that $ 75 $, $ 375 $ satisfies the condition, for the inductive step we prove that we can add an extra digit while still making it divisible by $5^{n + 1}$.

$ N_{n + 1} = a ( 10 )^n + N_{n} $
where $a$ is an odd number, and $5^n$ divides $10^n$ and $N_{n}$ from induction hypothesis.

so, after dividing by $5^n$, we still want the remaining number to be divisible by $5$, and notice that $a$ has $5$ odd options which is all the residue class mod $5$, so depending on $N_{n}$, such an $a$ must exist.

$\square$
:starwars:
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