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

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

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 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.

Monday, Jan 23rd

Problem
A rectangle can be divided into $n$ equal squares and also into $n+98$ equal squares. If the area is $n$, find its sides.

Solution
Let $x$ be the side of  each of the $n$ equal squares the rectangle can be divided into. Likewise define $y$ to be when it is divided into $n+98$ squares. Then  $nx^2=(n+98)y^2=n$ and $x=1$ and $y=\sqrt{\frac{n}{n+98}}$. Since there must be an integer number of $y$ squares on each row (column) and also and integer number of $x$ rectangles in each row (column), then $y$ has to be a rational number.

If $\text{gcd}(n,n+98)=1$, then $y^2$ is an irreducible fraction and both $n$ and $n+98$ have to be squares. Since $98\equiv 2 \text{ mod }4$, this is impossible.

Therefore $\text{gcd}(n,n+98)>1$, and hence $\text{gcd}(n,98)>1$. Let $n=dm$ where $d=\text{gcd}(n,98)$. Then $y^2=\frac{m}{m+98/d}$ and $\text{gcd}(m,m+98/d)=1$. Thus $m$ and $m+98/d$ have to be squares and that happens if and only if $2|d$. Let $m=p^2$ and $m+98/d=q$. If $d=2$, $49=(q-p)(q+p)$ and the only solutions are $q=25$ and $p=24$. If $d=14$, $7=(q-p)(q+p)$ with solutions $q=4$ and $p=3$.  If $d=98$, there are no solutions. Hence the possibilities for $n$ are $n=2\cdot 24^2$ and $n=14\cdot 3^2$.

Friday, Jan 20th

Problem
Julian writes down 5 positive integers such that their sum equals their product. Which numbers could have Julian writen down?

Solution
Let $a,b,c,d,e$ be the numbers Julian wrote down and without loss of generality suppose that $a\leq b\leq c\leq d\leq e$. If $a>1$, then $a+b+c+d+e\leq 5e$ and $abcde\geq 2^4 e$. Thus Julian must had written at least one 1. If $b>1$ something similar happens, as $a+b+c+d+e<5e$ and $abcde\geq 8e$. Hence $b=1$. If $c=1$ we have that $d=3$ and $e=3$ or $d=2$ and $e=5$ are solutions. If $c=2$, $d=2$ and $e=2$. If $c>2$, then $a+b+c+d+e<2e+5<4e$ and $abcde\geq 9e$, hence there are no solutions.

Therefore, the only solutions are $(1,1,1,2,5), (1,1,1,3,3), (1,1,2,2,2)$ and their permutations. 

Thursday, Jan 19th

Problem
How many 7-digit numbers are multiples of 388 and end in 388?

Solution
In order to a number $n$ be a multiple of 388 and to end in 388, we have to have that $n-388$ is also a multiple of 388, i.e. the problem reduces to find all 4 digit numbers $m$ such that $m10^3$ is a multiple of $388$.

Since $388=2^2\times 97$, $m$ has to be just a multiple of $97$. Thus $m=11\times97, 12\times 97\dots, 103\times 97$ which gives 93 possible numbers.

Wednesday, Jan 18th

Problem
Prove that if $11z^{10}+10i z^9+10i z-11=0$ then $|z|=1$.

Solution
Suppose that $|z|>1$. Then $|z^{9}(11z+10i)|>|z|^9>|10i z -11|$. Similarly, if $|z|<1$ we have that $|z^{9}(11z+10i)|<|z|^9<|10i z -11|$. Thus $|z|=1$.

Tuesday, Jan 17th

Problem
Find the least positive integer $n$ such that $19/(n+21), 20/(n+22), \dots, 91/(n+93)$ are all irreducible fractions.

Solution
For all these fractions to be irreducible, we must have that $gcd(18+m, n+20+m)=1$ for $m=1,2,\dots, 73$.  Therefore, $gcd(18+m, n+2)=1$ which gives that $n+2$ has to have a prime divisor bigger than 18+73, thus $n+2=97$ is the smaller solution and $n=95$ gives the answer.

Friday, Jan 13th

Problem
Sum the series $\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{m^2 n}{3^m(n3^m+m3^n)}$

Solution

Exchanging $m$ and $n$ produces the same series, so the sum will be
$\frac{1}{2}\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{m^2 n}{3^m(n3^m+m3^n)}+\frac{m n^2}{3^n(n3^m+m3^n)}=\frac{1}{2}\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{mn}{3^m 3^n}$

$=\frac{1}{2}\left(\sum_{k=1}^\infty \frac{k}{3^k}\right)^2=\frac{1}{2}\left(\frac{3}{4}\right)^2=\frac{9}{32}$

Thursday, Jan 12th

Problem
Let $x,y,z$ be positive reals such that $x^2+y^2+z^2=1$. Prove that $x^2yz+xy^2z+xyz^2\leq 1/3$

Solution
Notice that $x^2yz+xy^2z+xyz^2=xyz(x+y+z)$. Then using the AM-GM inequality together with the AM-RMS inequality gives the desired result.

Wednesday, Jan 11th

Problem
A 3 digit natural number is called tricubic if its the sum of the cubes of its digits. Find all pairs of consecutive tricubic numbers.

Solution
Suppose that $(n,n+1)$ is a pair of consecutive tricubic numbers and let $n=a10^2+b10+c$ be the decimal expansion of $n$. Then $n=a^3+b^3+c^3$ and $n+1=a^3+b^3+(c+1)^3$ (if $c\neq 9$) or $n+1=a^3+(b+1)^3$ (if $c=9$ and $b\neq 9$) or $n+1=(a+1)^3$ (if $c=b=9$).

In the first case, we have that $1=3c^2+3c+1$, from which $c=0$. Looking at the possibilities for $a$ and $b$ we find that there are no such tricubic numbers.

In the second case, we have that $9^3=3b(b+1)$, which has no solutions for $b$. In the third case, we have that $9^3+9^3=3a(a+1)$ which has no solutions for $a$.

Hence, there are no such pair of numbers.

Tuesday, January 10th

Problem

For each side of a polygon, divide its length by the length of the other sides. Prove that the sum of all such fractions is smaller than 2. 

Solution

With out loss of generality suppose that the sum of the sides $a_1,\dots, a_n$ of the polygon is 1. Hence it suffices to prove that

$\sum_{i=1}^n \frac{a_i}{1-a_i}\leq 2$

Since $f(x)=\frac{x}{1-x}$ is a convex function, we have that

$\sum_{i=1}^n \frac{a_i}{1-a_i}\leq \frac{n}{n-1}\leq 2$

Monday, January 9th

Problem
Determine if there exists $f:\mathbb{Z}^+\to \mathbb{Z}^+$ such that $f(f(n))=2n$ for all positive integers n.

Solution
As long as $f$ does not have a fixed point there is no problem in the definition of $f$. The key is to analyze the powers of 2, as it appears on the defining property of $f$.

We have that $f\left(2^p q\right)=2^p f(q)$ for $q$ an odd integer, hence $f$ is determined by the values it takes on the odd integers. Since the powers of 2 are not affected by $f$, it suffices that $f$ map $2^r a$ to $2^s b$, where $a,b$ are odd integers and $r,s$ are non-negative integers and the only restriction is that $a\neq b$.