First Name: Cruz

Last Name: Arnzen

UIN: $868755043$

Problem $1$

We know that if all three of the primes are odd then: \[

pq+qr+rp\equiv1+1+1\equiv1\pmod2,

\]

while

\[

p+q+r+2025\equiv1+1+1+1\equiv0\pmod2,

\]

which obviously doesn't make sense.

We know that every prime other than $2$ is congruent to $1\pmod{2}$.

We also know if all of them are odd then\[

pq+qr+rp \;\equiv\;1\cdot1 + 1\cdot1 + 1\cdot1 \;=\;3 \;\equiv\;1\pmod2,

\]

while

\[

p+q+r+2025 \;\equiv\;1+1+1 +\,2025\pmod2.

\]

Since $2025$ is odd one side would be odd while the other side is even which again doesn't make sense since we said earlier that this applies to all primes except $2$, we know $2$ is one of the primes. 

We can just say $p=2$.

Now we can just plug in giving us: $pq+qr+rp = 2q+qr+2r = q\,r+2(q+r),$ and

$p+q+r+2025 = 2+q+r+2025 = q+r+2027$. So

$qr+2(q+r)=q+r+2027

\quad\Longrightarrow\quad

qr+q+r=2027

\quad\Longrightarrow\quad

(q+1)(r+1)=2028.$

Since we know, $q$ and $r$ are primes we can use the prime factorization $2028=2^2\cdot 3\cdot 13^2$.

Now we can just check the divisors.

After checking, we see that the only way that they're both prime is if $q+1=6,\ r+1=338

\quad\Longrightarrow\quad

q=5,\ r=337.$

So we get our final answer of $$p+q+r=2+5+337=\boxed{344}.$$

​

First Name: Cruz

Last Name: Arnzen

UIN: $868755043$

Problem $2$

​

We can just factor it giving: $(mn-1 + m - n)\,(mn-1 - m + n)

\;=\;(mn-1)^2 \;-\;(m-n)^2

\;=\;2025\,.$

Now we can set $x = mn-1,\quad y = m-n,$

then

$x^2 - y^2 = 2025

\quad\Longrightarrow\quad

(x+y)(x-y)=2025.$

We know that the prime factorization of $2025=3^4\cdot5^2.$

We will  look at all of its odd divisors: $\pm1,\pm3,\pm5,\pm9,\pm15,\pm25,\pm27,\pm45,\pm75,\pm135,\pm225,\pm675,\pm2025.$

We can now use some clever algebra by using the fact that we know that $x+y=A,\quad x-y=B,\quad A\,B=2025.$ 

From this, we can easily get $x=\frac{A+B}2,\quad y=\frac{A-B}2$.

Now we can find $(m,n)$ by solving these equations: $mn = x+1,\quad m-n=y.$

By checking all divisors we find the solutions to be: $$(m,n)\;=\;

\bigl(\pm26,\pm2\bigr)

\quad\text{or}\quad

\bigl(\pm2,\pm26\bigr).$$

​

First Name: Cruz

Last Name: Arnzen

UIN: $868755043$

Problem $3$

​

Let $x = \text{number of correct answers},

y = \text{number of wrong answers}.$

Then $x+y\le90$ and the score is

$S \;=\;4x-2y \;=\;2\,(2x-y)\;=\;2t,$

where $t=2x-y.$

Our minimum is when $x=0,\;y=90:

t_{\min}=0-90=-90.$

Our maximum is when $x=90,\;y=0:

t_{\max}=180.$

So $t$ could be any integer in $[-90,180]$.

We see that every integer $x+y\le90$ in our range produces a positive solution, except $t=179\,\Longrightarrow S=2\cdot179=358$,

which needs one too many questions to be answered.

All of the even integers from $-180 \text{ to } 360$ inclusive:

$\frac{360-(-180)}2+1=271$ values.

Getting rid of the score gives us: $270$.

​

First Name: Cruz

Last Name: Arnzen

UIN: $868755043$

Problem $4$

​

We see that the discriminant is $D = 6^2 - 4\cdot n\cdot(-2025) = 36 + 8100\,n = 36\bigl(225n+1\bigr).$

​

We know that for $D$ to be a perfect square, $225n+1$ must be a perfect square. We can let $225n+1 = m^2,\quad m\in\mathbb{Z}^+

\;\Longrightarrow\;

n = \frac{m^2-1}{225}.$

Solving $(m^2\equiv1\pmod{225})$ gives the smallest normal solution $m=26$ plugging this in gives $n = \frac{26^2 - 1}{225} = \frac{675}{225} = 3.$ 

So the least positive integer $n$ is

$\boxed{3}.$ Number of integer $x$ with $f_3(x)<0$

For $n=3$, 

$f_3(x) = 3x^2 + 6x -2025 = 3\,(x-25)(x+27).$

So there are $51$ values between our bounds $[-25,27]$ , which is our answer.

​

First Name: Cruz

Last Name: Arnzen

UIN: $868755043$

Problem $5$

​

Since we know that $\text{MATH}$ is a four‑digit number,

$1000M \le MATH < 1000(M+1).$

If $M\ge2$, then $MATH\ge2000$, so

$CLE+VER+MATH\;\ge\; CLE+VER+2000\;\ge\;2000+ (100+100)\;=\;2200>2025,$

which is just impossible.

So $M=1$ and we see that $\text{MATH} = 1000 + D,\quad D=100A+10T+H.$

Now we can plug it in giving us: $\text{CLE + VER} + (1000 + \text{D}) = 2025

\quad\Longrightarrow\quad

\text{CLE + VER + D} = 1025.$

We can set $S=\text{CLE+VER}$ which is what we want to minimize.

We know that the two smallest integers available aside from $1$ are $2,3$ so we have $\text{CLE + VER} \;\ge\; 200+\dots +300+\dots \;=\;500+\dots$ we will use these as the leading digits.

With some trial and error we find $$CLE=250,\quad VER=307

\qquad\Longrightarrow\qquad

S = 250+307 = 557.$$

Subtracting gives us our final answer of $$1468.$$

​

​

First Name: Cruz

Last Name: Arnzen

UIN: $868755043$

Problem $6$

​

We can let $A = \sqrt{x^2 + 2025x - 1},\quad

B = \sqrt{x^2 + 25x - 1}.$ 

​

We know $A - B = 50$ so $A = B + 50.$

We can just square both sides $x^2 + 2025x -1 \;=\;(B+50)^2 = B^2 + 100B + 2500.$

Since we know $B^2 = x^2 + 25x -1,$

We see that $x^2 + 2025x -1 = x^2 + 25x -1 + 100B + 2500

\quad\Longrightarrow\quad

2000x = 100B + 2500

\quad\Longrightarrow\quad

B = 20x - 25.$

We can square again now $x^2 + 25x -1 = (20x - 25)^2

\;\Longrightarrow\;

399x^2 -1025x +626 =0.$

The discriminant is $\Delta =1025^2 -4\cdot399\cdot626 =227^2, so

x = \frac{1025\pm 227}{2\cdot399}

\;=\;\frac{1025+227}{798}= \frac{1252}{798}=\frac{626}{399}

\quad\text{or}\quad

\frac{1025-227}{798}=1.$

We will check both solutions now to make sure they are not extraneous. After we check both, we find that $x=1$ IS extraneous but $x=\dfrac{626}{399}$ is not.

So our solution is $x=\dfrac{626}{399}.$

​

First Name: Cruz

Last Name: Arnzen

UIN: $868755043$

Problem $7$

​

We know with Ptolemy’s theorem cyclic quadrilateral $ABCD$ diagonals satisfy

$AC\cdot BD = AB\cdot CD + BC\cdot DA = 2\cdot7 + 6\cdot9 = 14 + 54 = 68.$

We can use the fact that $\angle B + \angle D = 180^\circ$ in a cyclic quadrilateral, we see  see by the Law of Cosines in $\triangle ABC$ and $\triangle ADC$ that

$AC^2 = \frac{272}{5},$

so $AC = \sqrt{\frac{272}{5}} = \frac{4\sqrt{85}}{5}.$

So we know $BD = \frac{68}{AC} = \frac{68}{\frac{4\sqrt{85}}5}

= \sqrt{85}.$

We can see that $AC + BD = \frac{4\sqrt{85}}5 + \sqrt{85}

= \frac{9\sqrt{85}}5

= \frac{153}{\sqrt{85}}.$

We can now add our two values giving us $238.$

​

First Name: Cruz

Last Name: Arnzen

UIN: $868755043$

Problem $8$

​

We can $x = \sqrt{a + 8b},\quad

y = \sqrt{b + 8c},\quad

z = \sqrt{c + 8a}.$

We know that $x,y,z\ge0.$

\[

\begin{cases}

a + 8b = x^2,\\

b + 8c = y^2,\\

c + 8a = z^2.

\end{cases}

\]

Solving this linear system  gets us

$a = \frac{x^2 - 8y^2 + 64z^2}{513},\quad

b = \frac{64x^2 + y^2 - 8z^2}{513},\quad

c = \frac{-8x^2 + 64y^2 + z^2}{513}.$

We see $a + b + c \;=\;\frac{x^2+y^2+z^2}{9}.$

We can now rearrange the original equation giving $x + 4y + z

\;=\;\dfrac{x^2+y^2+z^2 + 18}{2}.$

Rearranging again gives us $x^2 -2x \;+\; y^2 -8y \;+\; z^2 -2z \;+\;18 \;=\;0.$

Now we can complete squares to give us $(x-1)^2 + (y-4)^2 + (z-1)^2 \;=\;0.$

This shows us that the only possible values are $x=1,\quad y=4,\quad z=1.$

Plugging it back in gives us $a=\frac{1 -128 +64}{513}=-\frac{7}{57},\quad

b=\frac{64 +16 -8}{513}=\frac{8}{57},\quad

c=\frac{-8 +1024 +1}{513}=\frac{113}{57}.$

We can check and all the radicals are real, so we're done!

​

First Name: Cruz

Last Name: Arnzen

UIN: $868755043$

Problem $9$

​

We know the equation

$m^3 + n^3 \;=\; 91\,m n \;-\;2025$

is symmetric in $m,n$.  If $m=n$, it becomes

$2m^3 \;=\;91m^2 \;-\;2025.$

From here, I'm not quite sure what to do, but we could probably rearrange and test values. 

​

​

First Name: Cruz

Last Name: Arnzen

UIN: $868755043$

Problem $10$

​

I believe we must use the Erdos Ginzburg Ziv theorem.