Problem:
Find a four digit number which is a perfect square, knowing that the first two digit number exceeds by $1$ the last two digit number.
Solution:
Let $x$ be the first two digit number and let $y^2$ be the four digit number. Then, $31 < y < 100$ and
$$100(x+1)+x=y^2 \iff 101x=(y-10)(y+10).$$ So, $101$ divides one between $y-10$ and $y+10$. But $y-10<101$, so $101|(y+10)$. Since $41<y+10<110$, $y+10=101$, which gives $y=91$. Therefore, $y^2=8281$.
Showing posts with label KöMaL. Show all posts
Showing posts with label KöMaL. Show all posts
Friday, December 21, 2012
Thursday, December 20, 2012
KöMaL (Metresis) 1894, Problem 10
Problem:
Find all integers such that their fifth power minus three times their square is equal to $216$.
Solution:
Suppose that $n$ is an integer such that $n^5-3n^2=216=2^3\cdot3^3$. Then $$n^2(n^3-3)=216.$$ From this equality we get $n>1$. Let $d=\gcd(n^2,n^3-3)$. Then $d=1,3$. If $d=1$, then both $n^2$ and $(n^3-3)$ are perfect cubes, impossible since $$(n-1)^3 < n^3-3 < n^3.$$ If $d=3$, then $3^2|n^2$, so $n^2=9a, n^3-3=3b$ and $\gcd(a,b)=1$. Therefore, $ab=8$ and $b+1=3^{3k-1}$ for some $k \in \mathbb{N}$, which gives $a=1,b=8$, i.e. $n=3$.
Note. It was faster to note that $n^2$ and $(n^3-3)$ have different parity, so one of the two is odd. If $n^2$ is odd, $n^2=9$ which yields $n=3$, if $(n^3-3)$ is odd, then $n^3-3=1,3,9,27$, i.e. no solution.
Find all integers such that their fifth power minus three times their square is equal to $216$.
Solution:
Suppose that $n$ is an integer such that $n^5-3n^2=216=2^3\cdot3^3$. Then $$n^2(n^3-3)=216.$$ From this equality we get $n>1$. Let $d=\gcd(n^2,n^3-3)$. Then $d=1,3$. If $d=1$, then both $n^2$ and $(n^3-3)$ are perfect cubes, impossible since $$(n-1)^3 < n^3-3 < n^3.$$ If $d=3$, then $3^2|n^2$, so $n^2=9a, n^3-3=3b$ and $\gcd(a,b)=1$. Therefore, $ab=8$ and $b+1=3^{3k-1}$ for some $k \in \mathbb{N}$, which gives $a=1,b=8$, i.e. $n=3$.
Note. It was faster to note that $n^2$ and $(n^3-3)$ have different parity, so one of the two is odd. If $n^2$ is odd, $n^2=9$ which yields $n=3$, if $(n^3-3)$ is odd, then $n^3-3=1,3,9,27$, i.e. no solution.
Sunday, October 14, 2012
KöMaL (Metresis) 1894, Problem 8
Problem:
If $2 \cos \vartheta = u+\dfrac{1}{u}$, prove that $2 \cos n \vartheta = u^n+\dfrac{1}{u^n}$ for every positive integer $n$.
Solution 1:
We prove the statement by induction on $n \geq 2$. If $n=2$ we have $$2 \cos 2\vartheta =4 \cos^2 \vartheta - 2=\left(u+\dfrac{1}{u} \right)^2 - 2=u^2+\dfrac{1}{u^2}.$$ Assume that the statement is true for every $2 \leq k \leq n$. Then, $$\begin{eqnarray*} 2 \cos (n+1)\vartheta &= &2[2\cos n \vartheta \cos \vartheta - \cos (n-1)\vartheta]\\ &= & 2\left[\left(u^n+\dfrac{1}{u^n}\right)\dfrac{1}{2}\left(u+\dfrac{1}{u}\right)-\dfrac{1}{2}\left(u^{n-1}+\dfrac{1}{u^{n-1}}\right)\right] \\ & = & u^{n+1}+\dfrac{1}{u^{n+1}}, \end{eqnarray*}$$ and the statement follows.
Solution 2:
From the given equality, we obtain $u^2-(2\cos \vartheta) u + 1=0$, and solving this equation in $u$ we get $u=\cos \vartheta \pm i \sin \vartheta=e^{\pm i \vartheta}$. Then, for every positive integer $n$
$$2 \cos n \vartheta = e^{\pm in\vartheta}+e^{\mp in \vartheta}=u^n+u^{-n}=u^n+\dfrac{1}{u^n}$$ as we wanted to prove.
If $2 \cos \vartheta = u+\dfrac{1}{u}$, prove that $2 \cos n \vartheta = u^n+\dfrac{1}{u^n}$ for every positive integer $n$.
Solution 1:
We prove the statement by induction on $n \geq 2$. If $n=2$ we have $$2 \cos 2\vartheta =4 \cos^2 \vartheta - 2=\left(u+\dfrac{1}{u} \right)^2 - 2=u^2+\dfrac{1}{u^2}.$$ Assume that the statement is true for every $2 \leq k \leq n$. Then, $$\begin{eqnarray*} 2 \cos (n+1)\vartheta &= &2[2\cos n \vartheta \cos \vartheta - \cos (n-1)\vartheta]\\ &= & 2\left[\left(u^n+\dfrac{1}{u^n}\right)\dfrac{1}{2}\left(u+\dfrac{1}{u}\right)-\dfrac{1}{2}\left(u^{n-1}+\dfrac{1}{u^{n-1}}\right)\right] \\ & = & u^{n+1}+\dfrac{1}{u^{n+1}}, \end{eqnarray*}$$ and the statement follows.
Solution 2:
From the given equality, we obtain $u^2-(2\cos \vartheta) u + 1=0$, and solving this equation in $u$ we get $u=\cos \vartheta \pm i \sin \vartheta=e^{\pm i \vartheta}$. Then, for every positive integer $n$
$$2 \cos n \vartheta = e^{\pm in\vartheta}+e^{\mp in \vartheta}=u^n+u^{-n}=u^n+\dfrac{1}{u^n}$$ as we wanted to prove.
KöMaL (Metresis) 1894, Problem 2
Problem:
Find all positive integers $N$ divisible only by $2$ and by $3$ such that $N^2$ has three times the number of divisors of $N$.
Solution:
Let $N=2^a 3^b$, where $a,b \in \mathbb{N}^*$. Then $N^2 = 2^{2a}3^{2b}$. So, it must be $$3(a+1)(b+1)=(2a+1)(2b+1) \iff ab-a-b-2=0 \iff (a-1)(b-1)=3.$$ Since $a-1 \geq 0$ and $b-1 \geq 0$, we obtain $$\left\{ \begin{array}{rcl} a-1 & = & 1 \\ b-1 & = & 3 \end{array} \right. \qquad \textrm{or} \qquad \left\{ \begin{array}{rcl} a-1 & = & 3 \\ b-1 & = & 1 \end{array} \right.$$ i.e. $a=2, b=4$ or $a=4, b=2$. Then all the positive integers required are $N_1=2^2 3^4=324$ and $N_2=2^4 3^2=144$.
Find all positive integers $N$ divisible only by $2$ and by $3$ such that $N^2$ has three times the number of divisors of $N$.
Solution:
Let $N=2^a 3^b$, where $a,b \in \mathbb{N}^*$. Then $N^2 = 2^{2a}3^{2b}$. So, it must be $$3(a+1)(b+1)=(2a+1)(2b+1) \iff ab-a-b-2=0 \iff (a-1)(b-1)=3.$$ Since $a-1 \geq 0$ and $b-1 \geq 0$, we obtain $$\left\{ \begin{array}{rcl} a-1 & = & 1 \\ b-1 & = & 3 \end{array} \right. \qquad \textrm{or} \qquad \left\{ \begin{array}{rcl} a-1 & = & 3 \\ b-1 & = & 1 \end{array} \right.$$ i.e. $a=2, b=4$ or $a=4, b=2$. Then all the positive integers required are $N_1=2^2 3^4=324$ and $N_2=2^4 3^2=144$.
KöMaL (Metresis) 1894, Problem 1
Problem:
Find all the two digit numbers $\overline{ab}$ such that $\overline{ababab}+1$ is a perfect cube.
Solution:
Let $x=\overline{ab}$ be the two digit number. Then, $101010x+1=n^3$, for some positive integer $n$. Clearly, $10^6 \leq n^3 < 10^7$, i.e. $100 \leq n \leq 215$. Moreover,
$$101010x=n^3-1.$$ Since $101010=30\cdot3367$, it's easy to see that $n^3-1 \equiv 0 \pmod{30}$ if and only if $n \equiv 1 \pmod{30}$, i.e. $n-1$ is divisible by $30$. So $n \in \{121,151,181,211\}$ and an easy check shows that only $n=211$ works. In this case, $x=93$.
Find all the two digit numbers $\overline{ab}$ such that $\overline{ababab}+1$ is a perfect cube.
Solution:
Let $x=\overline{ab}$ be the two digit number. Then, $101010x+1=n^3$, for some positive integer $n$. Clearly, $10^6 \leq n^3 < 10^7$, i.e. $100 \leq n \leq 215$. Moreover,
$$101010x=n^3-1.$$ Since $101010=30\cdot3367$, it's easy to see that $n^3-1 \equiv 0 \pmod{30}$ if and only if $n \equiv 1 \pmod{30}$, i.e. $n-1$ is divisible by $30$. So $n \in \{121,151,181,211\}$ and an easy check shows that only $n=211$ works. In this case, $x=93$.
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