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 $n=14\cdot 3^2$.
Tuesday, January 24, 2012
Sunday, January 22, 2012
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.
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.
Friday, January 20, 2012
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.
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.
Thursday, January 19, 2012
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 $|z|=1$.
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, January 17, 2012
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.
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
$\frac{1}{2}\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{m^2 n}{3^m(n3^m+m3^n)}+\frac{m n^2}{3^n(n3^m+m3^n)}=\frac{1}{2}\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{mn}{3^m 3^n}$
$=\frac{1}{2}\left(\sum_{k=1}^\infty \frac{k}{3^k}\right)^2=\frac{1}{2}\left(\frac{3}{4}\right)^2=\frac{9}{32}$
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
$\frac{1}{2}\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{m^2 n}{3^m(n3^m+m3^n)}+\frac{m n^2}{3^n(n3^m+m3^n)}=\frac{1}{2}\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{mn}{3^m 3^n}$
$=\frac{1}{2}\left(\sum_{k=1}^\infty \frac{k}{3^k}\right)^2=\frac{1}{2}\left(\frac{3}{4}\right)^2=\frac{9}{32}$
Friday, January 13, 2012
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.
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.
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