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.