Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "User:Ddk001"

m (→‎User Counts)
Line 3: Line 3:
 
If this is you first time visiting this page, please change the number below by one. (Add 1, do NOT subtract 1)
 
If this is you first time visiting this page, please change the number below by one. (Add 1, do NOT subtract 1)
  
−
<math>\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{0}}}}}}</math>
+
<math>\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{68}}}}}}</math>
  
 
Doesn't that look like a number on a pyramid?
 
Doesn't that look like a number on a pyramid?

Revision as of 14:45, 4 January 2024

User Counts

If this is you first time visiting this page, please change the number below by one. (Add 1, do NOT subtract 1)

$\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{68}}}}}}$

Doesn't that look like a number on a pyramid?

Cool asyptote graphs

Asymptote is fun! [asy]draw((0,0)----(0,6));draw((0,-3)----(-3,3));draw((3,0)----(-3,6));draw((6,-6)----(-6,3));draw((6,0)----(-6,0));[/asy]

[asy]draw(circle((0,0),1));draw((1,0)----(0,1));draw((1,0)----(0,2));draw((0,-1)----(0,2));draw(circle((0,3),2));draw(circle((0,4),3));draw(circle((0,5),4));draw(circle((0,2),1));draw((0,9)----(0,18));[/asy]

Problems I made

See if you can solve these:

1. (Much easier) There is one and only one perfect square in the form

$(p^2+1)(q^2+1)-((pq)^2-pq+1)$

where $p$ and $q$ are prime. Find that perfect square.

2. Suppose there is complex values $x_1, x_2,$ and $x_3$ that satisfy

$(x_i-\sqrt[3]{13})((x_i-\sqrt[3]{53})(x_i-\sqrt[3]{103})=\frac{1}{3}$

Find $x_{1}^3+x_{2}^3+x_{2}^3$.

3. Suppose

$x \equiv 2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6 \pmod{7!}$

Find the remainder when $\min{x}$ is divided by 1000.

4. Suppose $f(x)$ is a $10000000010$-degrees polynomial. The Fundamental Theorem of Algebra tells us that there are $10000000010$ roots, say $r_1, r_2, \dots, r_{10000000010}$. Suppose all integers $n$ ranging from $-1$ to $10000000008$ satisfies $f(n)=n$. Also, suppose that

$(2+r_1)(2+r_2) \dots (2+r_{10000000010})=m!$

for an integer $m$. If $p$ is the minimum possible positive integral value of

$(1+r_1)(1+r_2) \dots (1+r_{10000000010})$.

Find the number of factors of the prime $999999937$ in $p$.

5. (Much harder) $\Delta ABC$ is an isosceles triangle where $CB=CA$. Let the circumcircle of $\Delta ABC$ be $\Omega$. Then, there is a point $E$ and a point $D$ on circle $\Omega$ such that $AD$ and $AB$ trisects $\angle CAE$ and $BE<AE$, and point $D$ lies on minor arc $BC$. Point $F$ is chosen on segment $AD$ such that $CF$ is one of the altitudes of $\Delta ACD$. Ray $CF$ intersects $\Omega$ at point $G$ (not $C$) and is extended past $G$ to point $I$, and $IG=AC$. Point $H$ is also on $\Omega$ and $AH=GI<HB$. Let the perpendicular bisector of $BC$ and $AC$ intersect at $O$. Let $J$ be a point such that $OJ$ is both equal to $OA$ (in length) and is perpendicular to $IJ$ and $J$ is on the same side of $CI$ as $A$. Let $O’$ be the reflection of point $O$ over line $IJ$. There exist a circle $\Omega_1$ centered at $I$ and tangent to $\Omega$ at point $K$. $IO’$ intersect $\Omega_1$ at $L$. Now suppose $O’G$ intersects $\Omega$ at one distinct point, and $O’, G$, and $K$ are collinear. If $IG^2+IG \cdot GC=\frac{3}{4} IK^2 + \frac{3}{2} IK \cdot O’L + \frac{3}{4} O’L^2$, then $\frac{EH}{BH}$ can be expressed in the form $\frac{\sqrt{b}}{a} (\sqrt{c} + d)$, where $b$ and $c$ are not divisible by the squares of any prime. Find $a^2+b^2+c^2+d^2+abcd$.

Someone mind making a diagram for this?

Answer key & solution to the problems

I will leave a big gap below this sentence so you won't see the answers accidentally.






























dsf






fsd

Answer key

1. 049

2. 170

3. 736

4. 011

5. 054

Solutions

Problem 1

There is one and only one perfect square in the form

$(p^2+1)(q^2+1)-((pq)^2-pq+1)$

where $p$ and $q$ is prime. Find that perfect square.

Solution 1

$(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2 \cdot q^2 +p^2+q^2+1-p^2 \cdot q^2 +pq-1=p^2+q^2+pq$. Suppose $n^2=(p^2+1)(q^2+1)-((pq)^2-pq+1)$. Then, $n^2=(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2+q^2+pq=(p+q)^2-pq \implies pq=(p+q)^2-n^2=(p+q-n)(p+q+n)$, so since $n=\sqrt{p^2+q^2+pq}>\sqrt{p^2+q^2}$, $n>p,n>q$ so $p+q-n$ is less than both $p$ and $q$ and thus we have $p+q-n=1$ and $p+q+n=pq$. Adding them gives $2p+2q=pq+1$ so by Simon's Favorite Factoring Trick, $(p-2)(q-2)=3 \implies (p,q)=(3,5)$ in some order. Hence, $(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2+q^2+pq=\boxed{049}$.$\square$

Problem 2

Suppose there are complex values $x_1, x_2,$ and $x_3$ that satisfy

$(x_i-\sqrt[3]{13})((x_i-\sqrt[3]{53})(x_i-\sqrt[3]{103})=\frac{1}{3}$

Find $x_{1}^3+x_{2}^3+x_{2}^3$.

Solution 1

To make things easier, instead of saying $x_i$, we say $x$.

Now, we have $(x-\sqrt[3]{13})(x-\sqrt[3]{53})(x-\sqrt[3]{103})=\frac{1}{3}$. Expanding gives

$x^3-(\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103}) \cdot x^2+(\sqrt[3]{13 \cdot 53}+\sqrt[3]{13 \cdot 103}+\sqrt[3]{53 \cdot 103})x-(\sqrt[3]{13 \cdot 53 \cdot 103}+\frac{1}{3})=0$.

To make things even simpler, let $a=\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103}, b=\sqrt[3]{13 \cdot 53}+\sqrt[3]{13 \cdot 103}+\sqrt[3]{53 \cdot 103}, c=\sqrt[3]{13 \cdot 53 \cdot 103}+\frac{1}{3}$, so that $x^3-ax^2+bx-c=0$.

Then, if $P_n=x_{1}^n+x_{2}^n+x_{3}^n$, Newton's Sums gives

$P_1+(-a)=0$ $(1)$

$P_2+(-a) \cdot P_1+2 \cdot b=0$ $(2)$

$P_3+(-a) \cdot P_1+b \cdot P_1+3 \cdot (-c)=0$ $(3)$

Therefore,

$P_3=0-((-a) \cdot P_1+b \cdot P_1+3 \cdot (-c))$

$=a \cdot P_2-b \cdot P_1+3 \cdot c$

$=a(a \cdot P_1-2b)-b \cdot P_1 +3 \cdot c$

$=a(a^2-2b)-ab+3c$

$=a^3-3ab+3c$

Now, we plug in $a=\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103}, b=\sqrt[3]{13 \cdot 53}+\sqrt[3]{13 \cdot 103}+\sqrt[3]{53 \cdot 103}, c=\sqrt[3]{13 \cdot 53 \cdot 103}+\frac{1}{3}:$

$P_3=(\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103})^3-3(\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103})(\sqrt[3]{13 \cdot 53}+\sqrt[3]{13 \cdot 103}+\sqrt[3]{53 \cdot 103})+3(\sqrt[3]{13 \cdot 53 \cdot 103}+\frac{1}{3})$.

As we have done many times before, we substitute $x=\sqrt[3]{13},y=\sqrt[3]{53},z=\sqrt[3]{103}$ to get

$P_3=(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3(abc+\frac{1}{3})$

$=x^3+y^3+z^3+3x^2y+3y^2x+3x^2z+3z^2x+3z^2y+3y^2z+6xyz-3(x^2y+y^2x+x^2z+z^2x+z^2y+y^2z+3xyz)+3xyz+1$

$=x^3+y^3+z^3+3x^2y+3y^2x+3x^2z+3z^2x+3z^2y+3y^2z+6xyz-3x^2y-3y^2x-3x^2z-3z^2x-3z^2y-3y^2z-9xyz+3xyz+1$

$=x^3+y^3+z^3+1$

$=13+53+103+1$

$=\boxed{170}$. $\square$

Note: If you don't know Newton's Sums, you can also use Vieta's Formulas to bash.

Problem 3

Suppose

$x \equiv 2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6 \pmod{7!}$

Find the remainder when $\min{x}$ is divided by 1000.

Solution 1 (Euler's Totient Theorem)

We first simplify $\cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6:$

$2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6=42^4+6 \cdot 30^6=(\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)}$

so

$x \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)} \equiv 1 \pmod{5}$

$x \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv 0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv 0 \pmod{6}$

$x \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv 6 \cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)} \equiv 6 \pmod{7}$.

where the last step of all 3 congruences hold by the Euler's Totient Theorem. Hence,

$x \equiv 1 \pmod{5}$

$x \equiv 0 \pmod{6}$

$x \equiv 6 \pmod{7}$

Now, you can bash through solving linear congruences, but there is a smarter way. Notice that $5|x-6,6|x-6$, and $7|x-6$. Hence, $210|x-6$, so $x \equiv 6 \pmod{210}$. With this in mind, we proceed with finding $x \pmod{7!}$.

Notice that $7!=5040= \text{lcm}(144,210)$ and that $x \equiv 0 \pmod{144}$. Therefore, we obtain the system of congruences :

$x \equiv 6 \pmod{210}$

$x \equiv 0 \pmod{144}$.

Solving yields $x \equiv 2\boxed{736} \pmod{7!}$, and we're done. $\square$

Problem 4

Suppose $f(x)$ is a $10000000010$-degrees polynomial. The Fundamental Theorem of Algebra tells us that there are $10000000010$ roots, say $r_1, r_2, \dots, r_{10000000010}$. Suppose all integers $n$ ranging from $-1$ to $10000000008$ satisfies $f(n)=n$. Also, suppose that

$(2+r_1)(2+r_2) \dots (2+r_{10000000010})=m!$

for an integer $m$. If $p$ is the minimum possible positive integral value of

$(1+r_1)(1+r_2) \dots (1+r_{10000000010})$.

Find the number of factors of the prime $999999937$ in $p$.

Solution 1

Since all integers $n$ ranging from $-1$ to $10000000008$ satisfies $f(n)=n$, we have that all integers $n$ ranging from $-1$ to $10000000008$ satisfies $f(n)-n=0$, so by the Factor Theorem,

$n+1|f(n)-n, n|f(n)-n, \dots, n-10000000008|f(n)-n$

$\implies (n+1)n \dots (n-10000000008)|f(n)-n$.

$\implies f(n)=a(n+1)n \dots (n-10000000008)+n$

since $f(n)$ is a $10000000010$-degrees polynomial, and we let $a$ to be the leading coefficient of $f(n)$.

Also note that since $r_1, r_2, \dots, r_{10000000010}$ is the roots of $f(n)$, $f(n)=a(n-r_1)(n-r_2) \dots (n-r_{10000000010})$

Now, notice that

$m!=(2+r_1)(2+r_2) \dots (2+r_{10000000010})$

$=(-2-r_1)(-2-r_2) \dots (-2-r_{10000000010})$

$=\frac{f(-2)}{a}$

$=\frac{a(-1) \cdot (-2) \dots (-10000000010)-2}{a}$

$=\frac{10000000010! \cdot a-2}{a}$

$=10000000010!-\frac{2}{a}$

Similarly, we have

$(1+r_1)(1+r_2) \dots (1+r_{10000000010})=\frac{f(-1)}{a}=-\frac{1}{a}$

To minimize this, we minimize $m$. The minimum $m$ can get is when $m=10000000011$, in which case

$-\frac{2}{a}=10000000011!-10000000010!$

$=10000000011 \cdot 10000000010!-10000000010!$

$=10000000010 \cdot 10000000010!$

$\implies p=(1+r_1)(1+r_2) \dots (1+r_{10000000010})$

$=-\frac{1}{a}$

$=\frac{10000000010 \cdot 10000000010}{2}$

$=5000000005 \cdot 10000000010!$

, so there is $\left\lfloor \frac{10000000010}{999999937} \right\rfloor=\boxed{011}$ factors of $999999937$. $\square$

Problem 5

$\Delta ABC$ is an isosceles triangle where $CB=CA$. Let the circumcircle of $\Delta ABC$ be $\Omega$. Then, there is a point $E$ and a point $D$ on circle $\Omega$ such that $AD$ and $AB$ trisects $\angle CAE$ and $BE<AE$, and point $D$ lies on minor arc $BC$. Point $F$ is chosen on segment $AD$ such that $CF$ is one of the altitudes of $\Delta ACD$. Ray $CF$ intersects $\Omega$ at point $G$ (not $C$) and is extended past $G$ to point $I$, and $IG=AC$. Point $H$ is also on $\Omega$ and $AH=GI<HB$. Let the perpendicular bisector of $BC$ and $AC$ intersect at $O$. Let $J$ be a point such that $OJ$ is both equal to $OA$ (in length) and is perpendicular to $IJ$ and $J$ is on the same side of $CI$ as $A$. Let $O’$ be the reflection of point $O$ over line $IJ$. There exist a circle $\Omega_1$ centered at $I$ and tangent to $\Omega$ at point $K$. $IO’$ intersect $\Omega_1$ at $L$. Now suppose $O’G$ intersects $\Omega$ at one distinct point, and $O’, G$, and $K$ are collinear. If $IG^2+IG \cdot GC=\frac{3}{4} IK^2 + \frac{3}{2} IK \cdot O’L + \frac{3}{4} O’L^2$, then $\frac{EH}{BH}$ can be expressed in the form $\frac{\sqrt{b}}{a} (\sqrt{c} + d)$, where $b$ and $c$ are not divisible by the squares of any prime. Find $a^2+b^2+c^2+d^2+abcd$.

Someone mind making a diagram for this?

Solution 1

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=User:Ddk001&oldid=209851"