Problem MA 82 CRUX MATHEMATICORUM vol 46, no 7 pag 285

 

                    ANSWER CP : $\underset{\;\;n}max\;d_{n}=197$

 More precisely    $d_{197\cdot k -100}=197$ , $d_{n}=1\;\;if\;n\neq 197\cdot k-100$

Solution CP

      We apply the following properties:

(A)       $(a,b)=(a,-b)$;

(B)        $(a,b)=(a+b\cdot m,b)=(a,b+a \cdot m)$ , for any integers $m$;

(C)        if $(c,b)=1$ then $(a,b)=(a \cdot c,b)$.

Here $(a,b)$ denote "greatest common divisor" for integers $a$ and $b$. See for (C) an ancient post.

      Let $d_{n}=(a_{n},a_{n+1})=(n^{2}+2\cdot n+50, (n+1)^{2}+2(n+1)+50)=$

        $=(n^{2}+2\cdot n+50,n^{2}+4\cdot n+53)=$

        $\overset{(B)}=(n^{2}+2\cdot n+50,n^{2}-4\cdot n+53-(n^{2}+2\cdot n+50))=$

     $=(n^{2}+2\cdot n+50,2\cdot n+3)\;\overset{(C)}{=}\underset{(2,2n+3)=1}{=}{=}(2(n^{2}+2n+50),2n+3)=$

     $=((2n+3)n+n+100,2n+3)\overset{(B)}{=}(n+100,2n+3)\overset{(B)}{=}(n+100,2n+3-2(n+100))=$

     $\overset{(A)}{=}(n+100,197)$.

The number 197 being prime

             if $n+100=197\cdot k$ then $d_{n}=197$

             and if $n+100=197\cdot k+r$, $r=1, 2, ... ,196$ then $d_{n}=1$.

$\blacksquare$

          Examples

     $a_{96}=5458=2\cdot 4729;$

      $a_{97}=9653=7^{2}\cdot 197$,    hence $d_{96}=1$

      $a_{98}=9850=2\cdot 5^{2}\cdot 197 $,    hence $d_{97}=197$

      $a_{99}=10\; 049=13 \cdot 773$,     hence $d_{98}=1$

....................................................................................................

      $a_{293}=86\;485=5\cdot 7^{2}\cdot 353$

      $a_{294}=87\;074=2\cdot 13\cdot17\cdot 197$,     hence $d_{293}=1$

      $a_{295}=87\;665=5\cdot 89\cdot 197$,      hence $d_{294}=197$

      $a_{296}=88\;258=2\cdot 44\;129$,       hence $d_{295}=1$

A property of GCD (Greatest Common Divisor)

      We will denote the greatest common divisor of two integers $a$ and $b$ as $gcd(a,b)$. Some authors use $(a,b)$. A definition of $gcd(a,b)$, more generally valid in unitary commutative rings is

 

$d=gcd(a,b) \Leftrightarrow \begin{cases}d\;\mbox{divides}\;a\;\;and \;\;d\;\mbox{divides}\;b&(i)\\if\;d_{1}\;\mbox{divides} \;a\;and\;d_{1}\;\mbox{divides}\;b\;\;then\;\;d_{1}\;\mbox{divides}\;d&(ii)\end{cases}$ 

        Words $\underline{divides}$ means, in expresions "$d\;divides \;a$" etc, that

"there are an element $x$ in ring $\mathbb{Z}$ such that $d\cdot x=a$" etc.

We will denote "$d\;divides \;a$" by $d\mid a$.

 

     LEMMA If $gcd(c,b)=1$ then $gcd(a,b)=gcd(a\cdot c,b)$

 

Proof. Let $d=gcd(a,b)$; we have $d\mid a$ so $d \mid a\cdot c$. Thus, first

$d \mid a\cdot c$  and   $d \mid b  $  (i')

Secondly, let $\delta$ be such that

(1)                           $\delta \mid a\cdot c$  and  $\delta \mid b$.

$gcd(c,b)=1\Rightarrow$ there are integers $u$ and $v$ such that

(2)                                 $c\cdot u+b \cdot v =1$.

From $\delta \mid b \Rightarrow \delta \mid b\cdot v$  $\overset{(2)}{\Rightarrow} \delta \mid 1-c \cdot u \Rightarrow \delta \mid a-a\cdot c \cdot u \overset{(1)}{\Rightarrow} \delta \mid a$.

     Thus, (1) $\Rightarrow \delta \mid a$ and $\delta \mid b$ so that , via (ii) from definition,

$\delta \mid d$. Finally

$(1) \Rightarrow \delta \mid d$  (ii')

that means - from (i') and (ii"), $d=gcd(a,b)$.

 

 

$\blacksquare$

Remark. Here are another discussion about such phaenomena.

 

 

    

Problem MA78 - Crux Mathematicorum Vol. 46, No. 6 , June, 2020

 

 

 ANSWER CiP : $n=667$

 Verification: $T(667)+3 \cdot 667=6+6+7+3\cdot 667=19+2001=2020$

SOLUTION CiP

Let $n=\overline{abc}$, $a\neq 0$; the given equation is written

$$a+b+c+3\cdot (100a+10b+c)=2020$$

$$\Leftrightarrow a+b+c+30(10a+b)+3c=2020$$

where do we get

(1)$$30\cdot \overline{ab}=2020-a-b-4c.$$

From equation (1) we see that $30\mid2020-a-b-4c$ and because $a, b, c$  are digits in base ten, we have $a+b+4c\geq 9+9+36=54$. Then the right member of (1) is $\geq 1966$ so its values can only be 2010 or 1980. Also from 1 we get

(2)$$\overline{ab}=\frac{2020-a-b-4c}{30} .$$

If $a+b+4c=10$ then $\overline{ab}=\frac{2010}{30}=67$ but $a=6,b=7$ they cannot check the condition $a+b+4c=10$.

If $a+b+4c=40$ then $\overline{ab}=\frac{1980}{30}=66$ so $a=6,b=6$ and $c=\frac{40-6-6}{4}=7$ which is the answer.

$\blacksquare$

 ===============

Added February 8, 2021

             Good answer  see V47n01, page 8-9

              They write in their solution that $T(n)=2020-3n$ and since for a three-digit number $n$

$1 \leq T(n) \leq27$

$\Leftrightarrow \; 1 \leq 2020-3n \leq27$

.......$\Leftrightarrow \;\;665 \leq n \leq 673$.

           On the other hand $n \equiv T(n) \;(mod\;3)$ and $T(n)+3n=2020 \; \Rightarrow$

$\Rightarrow \;T(n) \equiv 2020\;(mod\;3)$ $\Rightarrow \;T(n) \equiv 1\;(mod\;3)\;\;\Rightarrow \;n \equiv 1 \; (mod\;3)\;....\;\Rightarrow \;3 \mid (n-1)$.

          From the above two observations, we know that, if such $n$ is possible then it must be either $667,\;670$ or $673$, and we check each possibility ....

= end added=

 

 

SIMULTANEOUS DIVISIBILITY of 1-FORMS

          Is about Problem 1 from Eőtvős Mathematical Competition 1894.

In Romanian there is this book: "Probleme de matematica pentru concurs";
the book can be downloaded from here (the solution is on the pages 23-28).

          The stattement of the problem can be written as a logical equivalence

$(1)$                                      $17 \; \mid 2x+3y \; \; \Leftrightarrow \; 17 \; \mid 9x+5y.$

          We will give a list of equivalences, "completely" in some sense, from which we can deduce (almost) any of such a relations. 

LEMA For any integers $x$, $y$ the following equivalences hold:

$17 \; \mid \; I(x,y)=2x+3y \; \Leftrightarrow \; 17 \; \mid \;II(x,y)=x-7x \; \Leftrightarrow \; 17 \; \mid \; III(x,y)=3x-4y \; \Leftrightarrow $

$\Leftrightarrow \;17 \; \mid \; IV(x,y)=5x-y \; \Leftarrow \; 17 \; \mid V(x,y)=7x+2y \; \Leftarrow \; 17 \; \mid \; VI(x,y)=8x-5y.$


Proof. We have $17 \; \mid 2x+3y \; \Leftrightarrow \; \widehat{2}x+\widehat{3}y=\widehat{0}$, where $\widehat{m}$ mean the congruence class modulo $17$ determined by $m$; their set is $\mathbb{Z}/17\mathbb{Z}$ (actually it is the field $GF(17)$). If we multiply last equation by $\widehat{9}$ and we take into account that, in $GF(17)$,

 $\widehat{9} \cdot \widehat{2}=\widehat{1}, \; \widehat{9} \cdot \widehat{3}=\widehat{10}=-\widehat{7}$

we obtain $\widehat{1}x-\widehat{7}y=\widehat{0} \; \Leftrightarrow \; 17 \; \mid \; x-7y$.
          In the same way, starting with $\widehat{2}x+\widehat{3}y=\widehat{0}$ and multiplyng it by $\widehat{10}$, $\widehat{11}$, $\widehat{12}$, $\widehat{4}$ respectively, we get the remained equivalences.

$\blacksquare$

          Apparently we didn't find the desired result. It comes if we were to multiply $\widehat{2}x+\widehat{3y}=\widehat{0}$ by $\widehat{13}$ and taking into account that

 $\widehat{13} \cdot \widehat{2}=\widehat{9}$ and $\widehat{13} \cdot \widehat{3}=\widehat{5}$ 

but patience $\cdot$ $\cdot$ $\cdot$ The word "completely" means that no other divisibility relation that can be obtained from $17 \mid 2x+3y$ it must be able to be expressed by ones of the forms $I-VI$ (and apliying other simple divisibility properties). From example
$$17 \; \mid I(x,y) \; \stackrel{LEMA} \Leftrightarrow  17 \; \mid VI(x,y) \Leftrightarrow 17 \; \mid 17x-VI(x,y)=9x+5y.$$

          Finally, we conclude anothers equivalences that we will use in a similar problem:

$7\; \mid x+2y \; \Leftrightarrow \; 7 \; \mid 2x-3y \; \Leftrightarrow \; 7 \; \mid 3x-y;$

$7\; \mid 3x+y \; \Leftrightarrow \; 7\; \mid x-2y \; \Leftrightarrow \; 7 \; \mid 2x+3y.$

TESTE pentru EVALUAREA NATIONALA 2020

---

Ministerul Educației și Cercetării a publicat un prim set de resurse de antrenament pentru elevii care vor susține examenele naționale [1]

Data publicării: 

Luni, 30 Martie, 2020

În contextul suspendării cursurilor, pentru pregătirea elevilor în vederea susținerii examenelor naționale, Ministerul Educației și Cercetării a publicat un prim set de resurse de antrenament pentru elevii care vor susține examenele naționale [2].

Resursele de antrenament tip test, realizate împreună cu Centrul Național de Evaluare și Examinare, respectă modelul subiectului de examen. Obiectivul acestora este acela de a exersa în această săptămână și de a continua efortul în vederea obținerii unor rezultate bune la viitoarele examene.

Baremele de evaluare și de notare vor fi publicate pe același site, vineri, 3 aprilie 2020. Rolul acestora este de autoevaluare și calibrare a modului în care au fost redactate răspunsurile.

Totodată, în cadrul emisiunilor Teleșcoala [3] vor fi prezentate unele soluții ale acestor teste de antrenament, cu explicațiile necesare pentru a înțelege cât mai bine răspunsurile din baremele de evaluare și de notare.

An Inequality in a Book - Sự bất bình đẳng trong một cuốn sách

You can read the book here

You can download it directly from here.
                                 
             Usually the first problem in a book is easier. Look at it

An algebraic solution I did not find despite the obvious geometric meaning.
Here is the picture with the solution in the book.

True, she is beautiful............

          
            We present below the geometrical meaning of the problem. 
    
           The equation 

(1)                                        $9x^{2}+8xy+7y^{2}=6$

 represent an ellipse; thanks to $GEOGEBRA$ she looks like this

You can see in the image above the point $A(\frac{1}{2};\frac{1}{2})$ that belongs to the ellipse (drawn a little clumsy). The center of this ellipse still remain origin $O(0;0)$ but the axes are the lines $y=\frac{\sqrt{17}-1}{4}x$ , $y=-\frac{\sqrt{17}+1}{4}x$ as will be seen in a later image.
         
         The inequality given in the statement $9a^{2}+8ab+7b^{2} \leqslant 6$ says about the point $M(a:b)$ it is inside the ellipse (1).

          The equation

(2)                                             $7x+5y+12xy=9$

reprezent a hyperbola with asymptotes $y=-\frac{7}{12}$ and $x=-\frac{5}{12}$, as shown below and point $A(\frac{1}{2};\frac{1}{2})$ also belongs to her.

           It is important that the point $A(\frac{1}{2};\frac{1}{2})$ belongs to both conics but crucial is that these conics are tangent at $A$  (and both at line $13x+11y=12$); see the image below

         Now, inequality $7a+5b+12ab \leqslant 9$ says that point $M(a;b)$ is outer the branches of hyperbola which is (geometrically) obvious.

$ \blacksquare$

         I hope that through these we have dismantled the mechanism that made this kind of problems to appear.
                                                      GOOD  LOCK !

TELEȘC8ALĂ PROBLEMA 2 (Temă) - pentru clasa a VIII-a

Lecția 5 PIRAMIDA PATRULATERĂ REGULATĂ

Vezi lecția aici.






     Enunțul  Problemei 2     Piramida patrulateră regulată $VABCD$ cu latura bazei 
                                        egală cu $18\;cm$ și ${ {\it \it A_{l}}}=540\;cm^{2}$. Determinați:
                                     a) volumul piramidei;
                                     b) tangenta unghiului format de o muchie laterală cu planul
                                          bazei;
                                     c) distanța de la centul bazei la o muchie laterală.

Răspuns CP : a) $1296\;cm^{3}$ ; b) $\frac{2\sqrt{2}}{3}$ ; c) $\frac{18\sqrt{34}}{17}\;cm$.

 Rezolvare CP

     a) Cu formula $A_{l}=\frac{P_{b}\cdot a_{p}}{2}$ și datele problemei obținem, înlocuind valorile date:

$540=\frac{4 \cdot 18 \cdot a_{p}}{2}$

sau $38 \cdot a{p}=540$ și rezultă $a_{p}=540:36=15$, deci

$VM=a_{p}=15 \;cm$.

În cazul pătratului $ABCD$ avem $OM=a_{b}=\frac{l_{4}}{2}=18:2$, deci

$OM=a_{b}=9\;cm$.

Din formula $h^{2}+a_{b}^{2}=a_{p}^{2}$ deducem $h^{2}=a_{p}^{2}-a_{b}^{2}=15^{2}-9^{2}=225-81=144$; atunci $h=\sqrt{144}$, adică
 

$VO=h=12\; cm$.

Calculăm ${V}=\frac{{A}_{b} \cdot h}{3}=\frac{l^{2} \cdot h}{3}=\frac{18^2 \cdot 12}{3}=1296 \;cm^{3}$.

       b) Din $proj_{ABC}\;V\;=O$ și $proj_{ABC}\;A=A$ rezultă că proiecția  segmentului $[VA]$ pe planul $(ABC)$ este $proj_{ABC}[VA]=OA$.

 Atunci unghiul format de muchia $VA$ cu planul bazei $(ABC)$ este $\angle VAO$.
Dar, în pătratul $ABCD$ avem $AO=R=R_{4}=\frac{l_{4}\sqrt{2}}{2}=\frac{18\sqrt{2}}{2}=9\sqrt{2}$. 
Rezultă $tg\widehat{ABC}=\frac{c_{o}}{c_{a}}=\frac {VO}{AO}=\frac{12}{9\sqrt{2}}=\frac{2\sqrt{2}}{3}$.

   c) 

     Construim în triunghiul dreptunghic$VAO$, cu unghiul drept în vârful $O$

$OE\; \perp\;VA$   $\Rightarrow$   $OE=d(O;VA)=\frac{c_{1}\cdot c_{2}}{ip}$    ca înalțime pe ipotenuză, 

deci calculând întâi

 $VA=\sqrt{VO^{2}+OA^{2}}=\sqrt{12^{2}+(9\sqrt{2})^{2}}=\sqrt{144+81\cdot 2}=\sqrt{306}=3\sqrt{34}$, obținem

$d(O;VA)=\frac{VO\cdot OA}{VA}=\frac{12\cdot 9\sqrt{2}}{\sqrt{306}}=\frac{18\sqrt{34}}{17}$.

$\blacksquare$

Ă Â Î Ș Ț  ă â î ș ț