Friday, September 28, 2012

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

Tuesday, September 25, 2012

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.

Sunday, September 23, 2012

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 $\sqrt{2}$.

Tuesday, September 18, 2012

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

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

Wednesday, June 20, 2012

Wednesday, June 20th

Problem
Find all positive integers that are 700 times the sum of its digits.

Solution
This statement is equivalent to the 700 replaced by 7. If $d$ is the number of digits, then the maximum value of 7 times the sum of digits is $63d$ and the minimum value of the number is $10^{d-1}$, hence the maximum value of $d$ is 3. Let $n=100a+10b+c$, where $a,b,c$ are digits, then
$100a+10b+c=7a+7b+7c$
and hence $93a+3b=6c$. 


Therefore $a=0$ and $b=2c$. The only posibilities for $b$ and $c$ are $(2,1)$, $(4,2)$, $(6,3)$, and $(8,4)$, thus the only numbers that satisfy the problem are $m=2100, 4200,6300$ and $8400$.




Wednesday, February 22, 2012

Wednesday, February 22nd

Problem
Prove that for every integer n the fraction $\frac{21n+4}{14n+3}$ cannot be reduced any further.

Solution
Since $3(14n+3)-2(21n+4)=1$ we have that $gcd(21n+4,14n+3)=1$ and then the fraction is reduced.