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.
Thursday, January 12, 2012
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.
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 10, 2012
Tuesday, January 10th
Problem
Solution
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.
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 9, 2012
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 $a\neq b$.
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 $a\neq b$.
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