Wednesday, September 26th

Problem

Find the minimum n for which 1999 can be written as the sum of n forth powers of positive integers.

Solution

Notice that

$1999=3\cdot5^4+3^4+2\cdot 2^4+11\cdot 1^4$ 

which gives the value of $n=17$. Brute force over the few cases shows that this is the minimum

Monday, September 24th

Problem

Prove that for at least 30% of the natural numbers $n$ from 1 to 1,000,000, $2^n$ starts with a 1.

Solution

Notice that if $2^{n}$ has one more digit in its decimal expansion than $2^{n-1}$, then it starts with a digit 1. Thus, it only suffices counting the digits on the decimal expansion of $2^n$ will give how many powers of 2 less than $2^n$ start with a 1. Therefore the density of such numbers is

$\frac{\log_{10} \left(2^n\right)}{n}=\log_{10} (2)\,,$

and since $2^{10}>10^3$, we have that $\log_{10} (2)>3/10$ and the result follows.

Friday, September 21st

Problem

Find all positive integers $(x,y)$ such that $x^3+y$ and $x+y^3$ are divisible by $x^2+y^2$.


Solution

Consider $x^3+y-x(x^2+y^2)=y-xy^2=y(1-xy)$. Notice that in order to $x^3+y$ and $x+y^3$ being divisible by $x^2+y^2$, $gcd(x,y)=1$ as otherwise there is a contradiction. Hence $x^2+y^2$ divides $1-xy$, but $|1-xy|\leq x^2+y^2$, thus $x=y=1$.

Wednesday, September 19th

Problem

The area of a trapezoid is 2 and the sum of its diagonals is 4. Find the height of the trapezoid.

Solution

Let $ABCD$ be the trapezoid, where $AB$ is parallel to $CD$. Notice that by increasing the length of $AB$ and decreasing $CD$ by the same amount, the conditions of the problem still hold (the diagonals $AD$ and $CB$ are being translated).  Thus, without loss of generality, suppose that $ABCD$ is a paralellogram, i.e. $AB=CD=2d$. Therefore $CB$ and $AD$ meet at their midpoint $M$. Hence the triangle $CMD$ has $CD=2d$ and $CM+MD=2$, hence by locating $C$ at $(-d,0)$ and $D$ at $(d,0)$ we have that $M$ satisfies the ellipse

$x^2+\frac{y^2}{1-d^2}=1$

together with the condition $(2d)(2y)=2$. Therefore 

$x^2=1-\frac{d^2y^2}{d^2(1-d^2)}=1-\frac{1/4}{d^2(1-d^2)}$,

and then $d^2(1-d^2)\geq 1/4$, but by AG-GM, $d^2(1-d^2)\leq 1/4$, thus $d=1/\sqrt{2}$, $x=0$, and $y=\frac{1}{\sqrt{2}}$ and the height of $ABCD$ is $\sqrt{2}$.

Monday, September 17th

Problem

In how many zeros does $2012!$ end?

Solution

Since each zero comes from multiplying a 2 with a 5, it suffices to count how many 5 factors are in $2012!$.  Hence the number of zeros is

$\sum_{k=1}^\infty \left\lfloor\frac{2012}{5^k}\right\rfloor=501$