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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 2025
0 replies
Inspired by 2024 Fall LMT Guts
sqing   2
N a minute ago by Jackson0423
Source: Own
Let $x$, $y$, $z$ are pairwise distinct real numbers satisfying $x^2+y =y^2 +z = z^2+x. $ Prove that
$$(x+y)(y+z)(z+x)=-1$$Let $x$, $y$, $z$ are pairwise distinct real numbers satisfying $x^2+2y =y^2 +2z = z^2+2x. $ Prove that
$$(x+y)(y+z)(z+x)=-8$$
2 replies
sqing
an hour ago
Jackson0423
a minute ago
Dividing Pairs
Jackson0423   2
N 17 minutes ago by Jackson0423
Source: Own
Let \( a \) and \( b \) be positive integers.
Suppose that \( a \) is a divisor of \( b^2 + 1 \) and \( b \) is a divisor of \( a^2 + 1 \).
Find all such pairs \( (a, b) \).
2 replies
Jackson0423
Apr 13, 2025
Jackson0423
17 minutes ago
Ellipse and Vectors
scls140511   1
N 26 minutes ago by Tigolf
Source: 2024 China Round 1 (Gao Lian)
7 Let $F_1$ and $F_2$ be the two foci of ellipse $\omega$. $P$ is a point on $\omega$. Let $O$ be the center of the excircle of $\triangle PF_1F_2$. When $\vec{PO} \cdot \vec{F_1F_2} = 2\vec{PF_1} \cdot \vec{PF_2}$, find the minimum eccentricity of $\omega$.
1 reply
1 viewing
scls140511
Sep 8, 2024
Tigolf
26 minutes ago
Maximum number of nice subsets
FireBreathers   1
N 27 minutes ago by FireBreathers
Given a set $M$ of natural numbers with $n$ elements with $n$ odd number. A nonempty subset $S$ of $M$ is called $nice$ if the product of the elements of $S$ divisible by the sum of the elements of $M$, but not by its square. It is known that the set $M$ itself is good. Determine the maximum number of $nice$ subsets (including $M$ itself).
1 reply
FireBreathers
Yesterday at 10:27 PM
FireBreathers
27 minutes ago
No more topics!
The last nonzero digit of factorials
Tintarn   4
N Apr 6, 2025 by Sadigly
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 2
For each integer $n \ge 2$ we consider the last digit different from zero in the decimal expansion of $n!$. The infinite sequence of these digits starts with $2,6,4,2,2$. Determine all digits which occur at least once in this sequence, and show that each of those digits occurs in fact infinitely often.
4 replies
Tintarn
Mar 17, 2025
Sadigly
Apr 6, 2025
The last nonzero digit of factorials
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Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 2
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Tintarn
9038 posts
#1
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For each integer $n \ge 2$ we consider the last digit different from zero in the decimal expansion of $n!$. The infinite sequence of these digits starts with $2,6,4,2,2$. Determine all digits which occur at least once in this sequence, and show that each of those digits occurs in fact infinitely often.
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AshAuktober
993 posts
#2
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This digit has to be even from $\nu_p$ analysis, and thus it suffices to look at the last digit mod t. Then we can verify the sequence repeats, and this gives (I think?) All even numbers.
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VideoCake
9 posts
#3
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Unusually difficult for problem 2, unless I missed something. Here is a quick sketch of my solution:
Basically you first show that \(\nu_2(n!) > \nu_5(n!)\) for every integer \(n \ge 2\) to reduce it down to the digits 2, 4, 6, 8. Now, using Legendre and the fact that \(1 \cdot 2 \cdot 3 \cdot 4 \equiv -1 \pmod 5\) one can show that if \(n = 5^{\ell}\) for some positive integer \(\ell\), then
\[\frac{n!}{10^{\nu_5(n!)}} \equiv 2^{\ell} \pmod 5\]which is enough, as \(2, 4, 6, 8\) all have different and non-zero residues modulo 5.
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MyobDoesMath
51 posts
#4
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My initial solution: Let $a_n$ be the given sequences. Use $\nu_2(n!)>\nu_5(n!)$ as above to show that all members of the sequence must be even.

Now as a motivation for the construction, one can see that in $16!$, $17!$, ... up to $24!$ all even digits appear once (you can find this if you simply start to calculate the first couple terms of the sequence). When $n$ is not divisible by $5$, it is easy to see that $a_n \equiv n \cdot a_{n-1} \mod 10$. However, dealing with $5 \mid n$ is not so easy. In the example above we have $a_{20} \equiv 2 a_{19}$ as the ending $0$ does not change anything.
This motivates us to look at $(100k+16)!$ up to $(100k+24)!$ where we have $a_n \equiv n \cdot a_{n-1} \mod 10$ for $n \not=100k+20$ and $a_{100k+20} \equiv 2 a_{100k+19}$ by the same argument as before. Here, we can compute all $a_n$ if we know $a_{100k+16}$, and we can see that all even numbers appear in the sequence regardless of which even digit $a_{100k+16}$ is, as we have one of the sequences (if I didn't miscalculate):

$2$, $4$, $2$, $8$, $6$, $6$, $2$, $6$, $4$
$4$, $8$, $4$, $6$, $2$, $2$, $4$, $2$, $8$
$6$, $2$, $6$, $4$, $8$, $8$, $6$, $8$, $2$
$8$, $6$, $8$, $2$, $4$, $4$, $8$, $4$, $6$

(And of course, it is no surprise that all even digits appear in the latter sequences if they all appear in the first. So maybe as an addition to the motivation, it is also clear that this works for all $k$ if it works for $k=0$.)

As there are infinitely many $k \in \mathbb{N}$, this shows that $2,4,6,8$ all appear infinitely often.

Remark
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Sadigly
147 posts
#5
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Claim: Odd numbers can't appear

Proof: Let $n!=\overline{Am\underbrace{000..0}_{k~times}}$ where $m$ is an odd number

$$\frac{n!}{10^k}=\overline{Am}$$
Since $\overline{Am}\in\{1;3;5;7;9\}~(mod ~10)$, we have $10\nmid\frac{5\times n!}{10^k}$, which implies $v_2(n!)\leq v_5(n!)$, which is absurd.


Proving the other one is a pain in the butt. Let $\lfloor n!\rfloor$ denote the nonzero part of $n!$. It should be obvious that $(n+1)\lfloor n!\rfloor\equiv\lfloor (n+1)!\rfloor~(mod~10)$ This works perfectly until $5\mid n+1$. In that case,since $5=\frac{10}2$, we can say $\lfloor (n+1)!\rfloor=\lfloor \frac{n!}{2}\rfloor$ (Keep in note that $\lfloor \overline{blahblah8}\rfloor\in\{4;9\}~(mod~10)$, but since $9$ is odd, it can't appear). Using these, with enough patience , one can find a loop
This post has been edited 2 times. Last edited by Sadigly, Apr 7, 2025, 12:45 PM
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