## 2187 in Star Wars

Posted: April 15, 2017 in Mathematics

Last night the Pearl was playing the Star Wars movie episode 4. When Princess Leia was captured in the Death Star, she was held in the cell number 2187. Cell “2187” sounds very familiar. The black dude who is casting as Finn in the Force Awaken also has the code name “FN-2187”. I don’t know if the number 2187 has any special meaning to the Star Wars story or the screenwriter, but 2187 reminds me of tutoring full time 10 years ago on topic like Geometric Sequence because $3^7=2187$.

There are also some interesting facts about the number 2187. If you write the number in backward, you get 7812. When you add the two numbers, you get 9999. When you split the number 2187 into two 2-digit numbers 21 and 87 and then multiple them, you get 1827. And also notice that $2187 = 3^7 = 3^3 \cdot 3^4 = 27 \cdot 81$.

## Stanford Summer Institute Math Problems

Posted: April 9, 2017 in Mathematics

(1)  Suppose $x, y \in\mathbb{Z}$.

$x^2-9y^2=(x-3y)(x+3y)=2017=1\cdot2017=2017\cdot1=(-1)(-2017)=(-2017)(-1)$

Case $(x-3y)(x+3y) = 1\cdot2017 \Rightarrow (x,y)=(1009, 336)$

Case $(x-3y)(x+3y) = 2017\cdot1 \Rightarrow (x,y)=(1009, -336)$

Case $(x-3y)(x+3y) = (-1)(-2017) \Rightarrow (x,y)=(-1009, -336)$

Case $(x-3y)(x+3y) = (-2017)(-1) \Rightarrow (x,y)=(-1009, 336)$

(2)  Since there are more 2’s than 5’s in 2017! and $2\cdot5=10$, all we need to do is to count the number of 5’s in the prime factorization of 2017!.

$\displaystyle \left\lfloor\frac{2017}{5}\right\rfloor + \left\lfloor\frac{2017}{25}\right\rfloor + \left\lfloor\frac{2017}{125}\right\rfloor + \left\lfloor\frac{2017}{625}\right\rfloor = 403 + 80 + 16 + 3 = 502$

(3)  Suppose that $n \in\mathbb{Z}$.

$n^2+9n+14 = m^2$, for some $m\in\mathbb{Z}$

By completing the square,

$\displaystyle \left(n+\frac{9}{2}\right)^2 - \frac{25}{4} = m^2$

$\displaystyle \left(n+\frac{9}{2}-m\right)\left(n+\frac{9}{2}+m\right) = \frac{25}{4}$

$(2n+9-2m)(2n+9+2m) = 25 = 1\cdot25 = 25\cdot1 = 5\cdot5 \\ = (-1)(-25) = (-25)(-1) = (-5)(-5)$

Case $1\cdot25 \text{ and } 25\cdot1 \Rightarrow 2n+9=\dfrac{1+25}{2}=13 \Rightarrow n=2$

Case $5 \cdot 5 \Rightarrow 2n+9=\dfrac{5+5}{2}=5 \Rightarrow n=-2$

Case $(-1)(-25) \text{ and } (-25)(-1) \Rightarrow 2n+9=-\dfrac{1+25}{2}=-13 \Rightarrow n=-11$

Case $(-5)(-5) \Rightarrow 2n+9=-\dfrac{5+5}{2}=-5 \Rightarrow n=-7$

If we define zero being a perfect square, then $n=2, -2, -11, -7$. Otherwise $n=2, -11$.

(4)  $ax^4+bx^3+cx^2+dx+e=0$

$x=\sqrt3+\sqrt5$

$x^2=8-2\sqrt{15}$

$x^3=18\sqrt3-14\sqrt5$

$x^4=124-32\sqrt{15}$

$a(124-32\sqrt{15}) + b(18\sqrt3-14\sqrt5) + c(8-2\sqrt{15}) + d(\sqrt3-\sqrt5) + e = 0$

$\sqrt{15}: \quad -32a-2c=0 \Rightarrow c=-16a \Rightarrow \dfrac{c}{a} = -16$

$\sqrt5: \quad -14b-d=0$

$\sqrt3: \quad 18b+d=0$

$\therefore b=d=0$

constant term: $\displaystyle 124a+8c+e=0 \Rightarrow 124+\frac{8b}{a}+\frac{c}{a}=0$

$\displaystyle \therefore \frac{e}{a} = -\frac{8c}{a}-124 = 128-124=4$

In particular,

$\displaystyle x^4+\frac{c}{a}x^2+\frac{e}{a}=0 \Rightarrow x^4-16x^2+4=0$

In general, for any integer n,

$a=n, b=0, c=-16n, d=0, e=4n$

(5)  Let $r$ be the radius of the smallest semi-circle.

$2\sqrt2 r + 4\sqrt2 r = 6\sqrt2 \quad \therefore \; r=1$

Note that the perimeter of semi-circle is $\pi r$,

$\text{perimeter of the curve} = \pi + 2\pi + 4\pi = 7\pi$

(6)  Let the side length of B, C, D, E, F, G, H, and I be b, c, d, e, f, g, h, and i respectively.

$g = f+1$

$b = f+2$

$c = f+3$

$d = f+3+e$

$i = f+3+2e$

$h = f+g = 2f+1$

$b+c+d=i+h \Rightarrow 3f+8+e = 3f+4+2e \Rightarrow e=4$

$d+i=b+g+h \Rightarrow 2f+18=4f+4 \Rightarrow f=7$

$\therefore \text{Area} = (d+i)(i+h) = 32\cdot33 = 1056$

(7)  By Fermat’s little theorem, $3^6 \equiv 1 \mod 7$

$\displaystyle 8\sum_{k=51}^{100}10^k + 10^{50}a + 9\sum_{k=0}^{49}10^k \pmod{7} \equiv \sum_{k=0}^{49}3^{k+51} + 3^{2}a + 2\sum_{k=0}^{49}3^k$

$\displaystyle \equiv -\sum_{k=0}^{49}3^k + 2a + 2\sum_{k=0}^{49}3^k \equiv 2a+\sum_{k=0}^{49}3^k \equiv 2a+\frac{3^{50}-1}{3-1} \equiv 0$

$\displaystyle 4a + 3^{50}-1 \equiv 0 \pmod{7}$

$\displaystyle 4a + 1 \equiv 0 \pmod7$

$\displaystyle a \equiv 5 \pmod7$

$\displaystyle \therefore a = 5$

(8)  By Pythagoras theorem,

$x^2+13 = 4^2 = 16 \therefore x=\sqrt3$

(9)  By Euclidean Algorithm,

$11n+7 = (7n+1)(1)+4n+6$

$7n+1 = (4n+6)(1) + 3n-5$

$4n+6 = (3n-5)(1) + n+11$

$3n-5 = (n+11)(3) - 38$

hence $(11n+7, 7n+1) = (n+11, -38)$

$(n+11, -38) > 1 \quad\text{if}\quad 2|n+11 \text{ or } 19|n+11$

$n+11\equiv 0 \mod 2 \Rightarrow n\equiv1 \mod 2 \quad\therefore n=2k+1$

$n+11\equiv 0 \mod 19 \Rightarrow n \equiv 8 \mod 19 \quad\therefore n=19k+8$

(10)  Divide the numbers from 1 to 100 into congruence classes mod 11.

[1] : 1, 12, … , 100; There are 10 members so we can choose at most 5 members, since we don’t want any pair of numbers with difference of 11. (Any adjacent members in the class have difference of 11)

[2] : 2, 13, … , 90; There are 9 members so we can choose at most 5 members.

[3] : 3, 14, … , 91; There are 9 members so we can choose at most 5 members.

… … , similarly for classes such as [4] to [11], each of them have 9 members. Therefore we can at most pick $5\times 11 = 55$ numbers without any pair in the set having difference of 11.