## 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, 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, 13, … , 90; There are 9 members so we can choose at most 5 members.

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

… … , similarly for classes such as  to , 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.

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