Taurus Investment Holdings, a global private equity real estate firm, has recently purchased a shallow-bay industrial portfolio in Minneapolis. This portfolio spans 636,051 square feet and is currently occupied by 96% with 71 tenants.
The acquisition of the Minneapolis Shallow Bay Portfolio adds to Taurus’ already extensive U.S. portfolio which now includes over 13.2 million square feet of industrial properties across the country. Recent acquisitions in both the Southwest and Northeast have contributed to this growth. The average building size within the Minneapolis Portfolio is approximately 40,000 square feet and it boasts an impressive average occupancy rate of 98%, demonstrating strong demand for shallow bay properties.
According to Taurus CEO Peter A Merrigan, this latest acquisition aligns with their strategy focused on last-mile industrial assets: “Given current market conditions and significant supply-demand imbalances, we see great potential for value creation within this portfolio.”
# Mock AIME I Problems/Problem II.
Retrieved from “https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_I_Problems/Problem_II&oldid=143557”
## Mathematical Forums
## Category: High School Olympiads
## Topic: Prove that there exists infinitely many positive integers n such that:
## Views: 207
## [enter: math-user1, num_posts=42, num_likes_received=6]
## [math-user1, num_likes=0]
Prove that there exists infinitely many positive integers $n$ such that:
$n\mid (3^n+4^n-7)$
[hide]I’m sorry if it’s too easy or well-known.[/hide]
Edit : Sorry guys! It was supposed to be $n\mid(3^{2^n}+4^{2^n}-7)$ instead!
Thanks @below!
@below : Thanks again! Edited 🙂
## [enter: math-user2, num_posts=701, num_likes_received=447]
## [math-user2, num_likes=0]
This is not true. Take $n$ to be a prime divisor of $3+4-7=-1$, which cannot exist.
However if it was supposed to be
\[n\mid 3^{2^n}+4^{2^n}-7\]then we can use the fact that for any odd number $a$, there exists an infinite number of primes dividing numbers in the form \[a^m+b^m-c,\qquad m>1,c=a+b.\]
# 2008 AMC 12B Problems/Problem 15.
The following problem is from both the 2008 AMC 12B #15 and also appeared as Problem B on the [[2020 AIME I]]:
## Contents.
1 Problem
2 Solution
2.1 Solution with Vieta’s Formulas (Official MAA)
2.2 Alternate solution using Vieta’s Formulas (by Richard Rusczyk)
3 Video Solutions
3.1 Video solution by AoPS – Official
3.2 Video solution by Aaron He – https://www.youtube.com/watch?v=y9lXoLcJvDY&t=19s
4 See Also
## Problem.
Let $\alpha_{}^{}$ and $\beta_{}^{}$ be roots of $x^{5}-6x-10$. Find $(\alpha_{}+\beta_{})^5+(\alpha_{ }-\beta_{ }) ^{5}$.
$\mathrm{(A)} \,-20 \qquad \mathrm{(B)}\,-10+\sqrt{21} \qquad \mathrm{(C)}Â {-10}\sqrt{21} Â (\textbf {D}) Â {-10} \qquad (\textbf {E})\, 0$
## Solution.
### Solution with Vieta’s Formulas (Official MAA).
Let $s = \alpha + \beta$ and $p = \alpha -\beta$. Then
\[s^5+p^5=(s+p)(s^4-s^{3}p+s^{2}p^{2}-sp^{3}+p^{4}).\]
By Vieta’s formulas,
\[6=s(\alpha ^{4}-{\alpha } ^{3}\beta +{\alpha } ^{2}{\beta } ^{2}-{\alpha }\,{ {\beta }} p({ {\,\,}}{{)}=\,( s-p) ( s{}^{}_{1. The sum of the roots is $\frac{-b}{a}$ where a is the coefficient of the leading term and b is the coefficient of second highest degree term in a polynomial. In this case it would be $\frac{-0}{1}=0$
The answer to our problem will be $(x+y)^5+(x-y)^5= x(x+y)^4+x(x-y)^4+ y(x+y)^y+ y(x-y)^(y)$
$x(16)+ x(256)+ y(-16)+ (-64)= 240x$
We can see that we have an extra factor of $15$, so we divide by fifteen to get our final answer as:
$\boxed{(D)-10}$.
~Solution by Richard Rusczyk
## Video Solutions.
### Video solution by AoPS – Official.
https://www.youtube.com/watch?v=HvJGfokrYQo
### Video solution by Aaron He – https://www.youtube.com/watch?v=y9lXoLcJvDY&t=19s.
https://www.youtube.com/watch?v=y9lXoLcJvDY&t=19s
## Mathematical Forums
## Category: High School Olympiads
## Topic: A problem of divisibility with prime numbers
## Views: 138
## [enter: math-user1, num_posts=18, num_likes_received=0]
## [math-user1, num_likes=0]
$p,q$ are two distinct odd prime numbers. Prove that $p^{q-1}+q^{p-1}$ is divisible by $pq$
I have tried to use Fermat’s little theorem and Wilson’s theorem but I cannot solve it.
Can anyone help me? Thanks a lot!
[color=#FF0000][Mod edit: Do not double post http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=445381][/color]x^2 + (y-k)^2 = r^2
This equation represents a circle centered at the point (0,k) with radius r. The variable x represents the horizontal distance from the center and y represents the vertical distance from the center. The value k shifts or translates the circle up or down on a coordinate plane depending on whether it is positive or negative. Changing values of k will shift where exactly in relation to other objects this particular circle would be located.
# Problem
Given:
a = 5,
b = 6,
c = -3,
Find:
(a+b)/(-3)
Solution:
Substitute given values into expression:
(5+6)/(-3)
Simplify numerator first then denominator next:
11/-3=-11/3 ANSWER:(a+b)/(-3)=-11/4 ANSWEr:-33/12 OR -2.75## Mathematical Forums
## Category: High School Olympiads
## Topic: Inequality em717
## Views: 170
## [enter: math-user1, num_posts=3424, num_likes_received=351]
## [math-user1, num_likes=3]
Let $a,b,c>0$. Prove that:
\[\frac{ab}{(b+c)(c+a)}+\frac{bc}{(c+a)(a+b)}+\frac{ca}{(a+b)(b+c)}\le\left(\sqrt{\frac{(ab+bc+ca)^2-abc(a+b+c)}{(a^2+b^2+c^2)^3}}-\sqrt{\frac{9(ab+bc+ca)abc(a-b-c)^6}{8(a^4(b-c)+b^4(c-a)+c^4(a-b))^5}}\right)\cdot abc \]# Problem
Given a number of 10000000000 (10 billion), find the largest integer n that is not greater than the square root of this number.
The largest integer n is equal to sqrt(n) = sqrt(10000000000)=3162277.660168379331998893544432718533719555139325216825250.
So we can say that the largest integer n is equal to 3162277.Solve for x:
x + (x/5) = 18
To solve for x in this equation, we need to isolate it on one side of the equation by performing inverse operations.
Firstly, let’s combine like terms on the left side by adding x and (x/5). This gives us:
(x + x/5) = ((6x)/5)
Next step would be multiplying both sides by “the reciprocal” or “multiplicative inverse” which means that we need to multiply both sides by 5/6. This will cancel out the fraction on the left side, and give us:
(5/6)(x + x/5) = (18)(5/6)
Simplifying this gives us:
(x + x) = 15
2x = 90
Finally, divide both sides by 2 to isolate x.
2x / (2) = (90)/(2)
This leaves us with our final answer of:
[ b]X=45[/b]
## Mathematical Forums
## Category: High School Olympiads
## Topic: Find all pairs of positive integers a,b satisfying:
## Views: 162
## [enter: math-user1, num_posts=24, num_likes_received=4]
## [math-user1, num_likes=0]
Find all pairs $(a,b)$ of positive integers such that $ab+3$ divides $a^3+b$
[color=#FF0000][Mod : See http://www.artofproblemsolving.com/Wiki/index.php/LaTeX .][/color]
[color=#008000]Moderator says:[quote=”math-user1″]…and I don’t know how to use LaTeX.[/quote]Then learn it! It’s not hard at all![url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX%3AAbout][u][color=#800080]Here[/color][/u][/url]’s a good place for beginners. ;)[size=85](And please do not double post.)[/size][/color]
[size=85](I deleted your other topic in Pre-Olympiad as you posted it twice there too.) [/size]
[url=http://www.artofproblemsolving.com/blog/edit-entry.php?t=388556&ref_id=&l=a&p=-999#comments][b]* * *[/b][/url] [color=#FF0000][Mod : See http://www.artofproblemsolving.com/Wiki/index.php/LaTeX .][/color]
[color=#008000]Moderator says:[quote=”math-user1″]…and I don’t know how to use LaTeX.[/quote]Then learn it! It’s not hard at all![url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX%3AAbout][u][color=#800080]Here[/color][/u][/url]’s a good place for beginners. ;)[size=85](And please do not double post.)[/size]
(I deleted your other topic in Pre-Olympiad as you posted it twice there too.) [/color]
## [enter: math-user2, num_posts=256, num_likes_received=72]
## [math-user2, num_likes=0]
Let $d=\gcd(a^3+b, ab
