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24th PMO, Qualifying Stage #7
elpianista227   1
N an hour ago by elpianista227
Suppose $a, b, c$ are the roots of the polynomial $x^3 + 2x^2 + 2$. Let $f$ be the unique monic polynomial whose roots are $a^2, b^2, c^2$. Find $f(1)$.
1 reply
elpianista227
an hour ago
elpianista227
an hour ago
27th Philippine Mathematical Olympiad Area Stage #5
Siopao_Enjoyer   1
N an hour ago by Siopao_Enjoyer
Find the sum of the cubes of the roots of the polynomial p(x)=x^3-x^2+2x-3.
1 reply
Siopao_Enjoyer
an hour ago
Siopao_Enjoyer
an hour ago
Help me please
dssdgeww   1
N an hour ago by whwlqkd
Prove that there exists a positive integer n with 2024 prime divisors such that n| 2^n + 1
1 reply
dssdgeww
3 hours ago
whwlqkd
an hour ago
[PMO20 Qualifying I.13] Log raised to Log
LilKirb   1
N 3 hours ago by LilKirb
Find the sum of the solutions to the logarithmic equation:
\[ x^{\log{x}} = 10^{2 - 3\log{x} + 2(\log{x})^2}\]where $\log{x}$ is the logarithm of $x$ to the base $10$
1 reply
LilKirb
3 hours ago
LilKirb
3 hours ago
my brain isn't working :(
missmaialee   41
N 3 hours ago by sl1345961
Compute $(-1)^{11}-1^{10}+2^9+(-2)^8$.
41 replies
missmaialee
Yesterday at 9:45 PM
sl1345961
3 hours ago
Inequalitis
sqing   0
4 hours ago
Let $ a,b,c\geq  0 , a^2+b^2+c^2 =3.$ Prove that
$$a^3 +b^3 +c^3 +\frac{11}{5}abc  \leq \frac{26}{5}$$
0 replies
sqing
4 hours ago
0 replies
Algebraic Manipulation
Darealzolt   4
N 4 hours ago by jasperE3
It is known that \(a,b \in \mathbb{R}\) that satisfies
\[
a^3+b^3=1957
\]\[
(a+b)(a+1)(b+1)=2014
\]Hence, find the value of \(a+b\)
4 replies
Darealzolt
Yesterday at 4:01 AM
jasperE3
4 hours ago
not obvious trig identity!
mathmax001   1
N 4 hours ago by joeym2011
Problem ( trigonometry )
Let $ x \in \mathbb{R} $ and n a positive integer $ n >=1 $, Show that : $$ \tan\left({\frac{(n+1)x}{2}}\right)= \frac{\sum_{k=1}^n{\sin kx}}{\sum_{k=1}^n{\cos kx}} $$
Here is my take in this video: https://youtu.be/DBPyHNqk0GI?si=9r-YDuwv794AGe1p
1 reply
mathmax001
5 hours ago
joeym2011
4 hours ago
confused
greenplanet2050   3
N Today at 12:19 AM by mathprodigy2011
um something weird happened today

I was doing the 2002 aime ii and i tried #9

I used PIE with $(2^{10}-1)-(\text{Number of times there are n same elements})$

so for like 1 same element i did $2^9 \cdot \dbinom{10}{1}$ cause there are 10 ways to choose 1 element that will be repeated. Similarly for 2 same elements it would be $2^8 \cdot \dbinom{10}{2}$

So if $A_n=2^{10-n} \cdot \dbinom{10}{n},$ the answer would be $(2^{10}-1)-([A_1+A_3+A_5+A_7+A_9]-[A_2+A_4+A_6+A_8+A_{10}].$ But this number turned out to be $0.$

Later when looking at the solution, i found out that the correct number was $28501.$ But I realized that $A_2+A_4+A_6+A_8+A_{10}=28501.$ So I was really confused of why i got the right answer somehow in my calculations.

Can someone explain why this happened? Thanks! :)
3 replies
greenplanet2050
Yesterday at 6:29 PM
mathprodigy2011
Today at 12:19 AM
Easy one
irregular22104   2
N Yesterday at 10:27 PM by trangbui
Given two positive integers $a,b$ written on the board. We apply the following rule: At each step, we will add all the numbers that are the sum of the two numbers on the board so that the sum does not appear on the board. For example, if the two initial numbers are $2,5$; then the numbers on the board after step 1 are $2,5,7$; after step 2 are $2,5,7,9,12;...$
1) With $a = 3$; $b = 12$, prove that the number 2024 cannot appear on the board.
2) With $a = 2$; $b = 34$, prove that the number 2024 can appear on the board.
2 replies
irregular22104
May 6, 2025
trangbui
Yesterday at 10:27 PM
a