Join our FREE webinar on May 1st to learn about managing anxiety.

G
Topic
First Poster
Last Poster
Frankenstein FE
NamelyOrange   3
N 3 hours ago by jasperE3
[quote = My own problem]Solve the FE $f(x)+f(-x)=2f(x^2)$ over $\mathbb{R}$. Ignore "pathological" solutions.[/quote]

How do I solve this? I made this while messing around, and I have no clue as to what to do...
3 replies
NamelyOrange
Jul 19, 2024
jasperE3
3 hours ago
Find the domain and range of $f(x)=\frac{1}{1-2\cos x}.$
Vulch   1
N 5 hours ago by aidan0626
Find the domain and range of $f(x)=\frac{1}{1-2\cos x}.$
1 reply
Vulch
5 hours ago
aidan0626
5 hours ago
Find the domain and range of $f(x)=\frac{4-x}{x-4}.$
Vulch   1
N 5 hours ago by aidan0626
Find the domain and range of $f(x)=\frac{4-x}{x-4}.$
1 reply
Vulch
5 hours ago
aidan0626
5 hours ago
Find the domain and range of $f(x)=2-|x-5|.$
Vulch   0
5 hours ago
Find the domain and range of $f(x)=2-|x-5|.$
0 replies
Vulch
5 hours ago
0 replies
Inequalities
sqing   5
N Today at 1:13 AM by pooh123
Let $a,b,c> 0$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1.$ Prove that
$$  (1-abc) (1-a)(1-b)(1-c)  \ge 208 $$$$ (1+abc) (1-a)(1-b)(1-c)  \le -224 $$$$(1+a^2b^2c^2) (1-a)(1-b)(1-c)  \le -5840 $$
5 replies
sqing
Jul 12, 2024
pooh123
Today at 1:13 AM
Amc 10 mock
Mathsboy100   4
N Today at 1:01 AM by pooh123
let \[\lfloor  x   \rfloor\]denote the greatest integer less than or equal to x . What is the sum of the squares of the real numbers x for which \[  x^2 - 20\lfloor x \rfloor + 19 = 0  \]
4 replies
Mathsboy100
Oct 9, 2024
pooh123
Today at 1:01 AM
Inequalities
sqing   0
Today at 12:20 AM
Let $ a,b,c>0 . $ Prove that
$$ \left(1 +\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right )\geq 4\left(\frac{a+b}{b+c}+ \frac{b+c}{a+b}\right)$$$$ \left(1 +\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right )\geq \frac{32}{9}\left(\frac{a+b}{b+c}+ \frac{c+a}{a+b}\right)$$$$ \left(1 +\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right )\geq  \frac{8}{3}\left(  \frac{a+b}{b+c}+ \frac{b+c}{c+a}+ \frac{c+a}{a+b}\right)$$$$ \left(1 +\frac{a^2}{b^2}\right)\left(1+\frac{b^2}{c^2}\right)\left(1+\frac{c^2}{a^2}\right )\geq \frac{8}{3}\left(  \frac{a^2+bc}{b^2+ca}+\frac{b^2+ca}{c^2+ab}+\frac{c^2+ab}{a^2+bc}\right)$$
0 replies
sqing
Today at 12:20 AM
0 replies
Purple Comet High School Math Meet 2024 P1
franklin2013   3
N Today at 12:10 AM by codegirl2013
Joe ate one half of a fifth of a pizza. Gale ate one third of a quarter of that pizza. The difference in the amounts that the two ate was $\frac{1}{n}$ of the pizza, where $n$ is a positive integer. Find $n$.
3 replies
franklin2013
Yesterday at 11:20 PM
codegirl2013
Today at 12:10 AM
Cool vieta sum
Kempu33334   0
Yesterday at 11:44 PM
Let the roots of \[\mathcal{P}(x) = x^{108}+x^{102}+x^{96}+2x^{54}+3x^{36}+4x^{24}+5x^{18}+6\]be $r_1, r_2, \dots, r_{108}$. Find \[\dfrac{r_1^6+r_2^6+\dots+r_{108}^6}{r_1^6r_2^6+r_1^6r_3^6+\dots+r_{107}^6r_{108}^6}\]without Newton Sums.
0 replies
Kempu33334
Yesterday at 11:44 PM
0 replies
Inequality, tougher than it looks
tom-nowy   2
N Yesterday at 11:28 PM by pooh123
Prove that for $a,b \in \mathbb{R}$
$$ 2(a^2+1)(b^2+1) \geq 3(a+b). $$Is there an elegant way to prove this?
2 replies
tom-nowy
Apr 29, 2025
pooh123
Yesterday at 11:28 PM
2024 8's
Marius_Avion_De_Vanatoare   3
N Apr 3, 2025 by EVKV
Source: Moldova JTST 2024 P2
Prove that the number $ \underbrace{88\dots8}_\text{2024\; \textrm{times}}$ is divisible by 2024.
3 replies
Marius_Avion_De_Vanatoare
Jun 10, 2024
EVKV
Apr 3, 2025
2024 8's
G H J
G H BBookmark kLocked kLocked NReply
Source: Moldova JTST 2024 P2
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Marius_Avion_De_Vanatoare
55 posts
#1
Y by
Prove that the number $ \underbrace{88\dots8}_\text{2024\; \textrm{times}}$ is divisible by 2024.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MathLuis
1517 posts
#2 • 1 Y
Y by megarnie
We basically want to prove $2024 \mid 8 \cdot \frac{10^{2024}-1}{9}$, or just $2277 \mid 10^{2024}-1$.
Now since $2277=3^2 \cdot 11 \cdot 23$ we have $\phi(2277)=\phi(3^2) \cdot \phi(11) \cdot \phi(23)=6 \cdot 10 \cdot 22=1320$, now since $\gcd(2024, 1320)=88$ we must prove $2277 \mid 10^{88}-1$. Now from here we will split the division in prime powers which makes sense due to crt.
$3^2 \mid 10^{22k}-1, 11 \mid 10^{22k}-1, 23 \mid 10^{22k}-1$ all hold trivially, the last one from fermat's theorem, therefore we have $2277 \mid 10^{22k}-1$ for all positive integers $k$ and ofc when $k=4$ we indeed get $2277 \mid 10^{88}-1 \mid 10^{2024}-1$ which implies the original problem thus we are done :cool:.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
aidan0626
1882 posts
#3
Y by
We have $2024=8\cdot11\cdot23$. Clearly, the number is divisible by 8, and we can see that it is also divisible by 11 from the divisibility rule. Now, all we have to show is that it is divisible by 23 now, or that $23\mid \frac{8}{9}(10^{2024}-1)$. Since both 8 and 9 are relatively prime to 23, this is the same as showing that $23\mid 10^{2024}-1.$ This is equivalent to $10^{2024}\equiv1\pmod{23}$, which is true by Fermat's.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
EVKV
70 posts
#4
Y by
The number is equivalent to $8(\frac{10^{2024}-1}{9})$
$2024=11$x$23$x$2^{2}$
clearly 4 divides it
$10\equiv 1$ $mod$ $11$
so 11 divides it too
$8(\frac{10^{2024}-1}{9})$ $\equiv$ $8(\frac{(10^{22})^{92}-1}{9})$ $\equiv$ $8(\frac{(1)^{92}-1}{9})$ $\equiv$ $1$ $mod$ $23$

as $gcd(9,2024) = 1$ it has a valid inverse allowing this
This post has been edited 1 time. Last edited by EVKV, Apr 3, 2025, 2:05 AM
Z K Y
N Quick Reply
G
H
=
a