Karmaşık görünen 29 071 numaralı sorunun çok basit olduğu ortaya çıktı // The Problem 29 071 that seemed complicated turned out to be too trivial

The author, Mihály Bencze, is famous, which is perhaps why the Magazine GM-B accepted this issue.

In tranlation

                          "29071.   Let the function  $f\;:\;\mathbb{N} \to \mathbb{N}$  be defined by

$f(n)=\left [ \frac{n}{3}\right ]+\left [ \frac{n+1}{5} \right ]+\left [ \frac{n+2}{7}\right ]$ ,  for all $n\in\mathbb{N}.$

                                  Prove that the function is neither injective nor surjective."

ANSWER CiP

$f(0)=f(1)=0$  , so the function is NOT injective;

$f(8)=4,f(9)=6$ and the function $f$, which is increasing, never takes the value $5.$

                Solution CiP

               It was not said in the statement, as many other times, that  $[\alpha]$  denotes the integer part of the real number  $\alpha$. By definition, $[\alpha]\in\mathbb{Z}$  is the (uniquely determined) integer that satisfies one of the conditions below:

$$[\alpha]\leqslant \alpha <[\alpha]+1,\quad \quad \alpha -1<[\alpha]\leqslant \alpha\;. \tag{][}$$

 

          The case of injectivity:

$f(0)=\left [ \frac{0}{3}\right ]+\left [ \frac{1}{5}\right ]+\left [\frac{2}{7}\right ]=0+0+0=0:$

$f(1)=\left [\frac{1}{3}\right ]+\left [\frac{2}{5}\right ]+\left [\frac{3}{7}\right ]=0+0+0=0.$

So the function $f$ is NOT injective.

          The case of surjectivity:

     Let's first note that $f$ is increasing, because function $[x]$ is: $x<y\Rightarrow [x]\leqslant [y]$

        (The pedantic solver would do this:

        let $x<y;$ if $x\leqslant [y]$ then, because $[x]\leqslant x$ we deduce from the transitivity of the inequalities that $[x]\leqslant [y];$

                           if $[y]<x$ then, because $x<y\overset{(][)}{<}[y]+1$ we deduce from transitivity that $[y]<x<[y]+1\overset{(][)}{\Rightarrow} [x]=[y]$.)

          We see that $f(8)=\left [2\frac{2}{3}\right ]+\left [1\frac{4}{5}\right ]+\left [1\frac{3}{7}\right ]=2+1+1=4$,

$f(9)=\left [\frac{9}{3}\right ]+\left [\frac{10}{5} \right ]+\left [1\frac{4}{7}\right ]=3+2+1=6$

so $f(n)$ takes the values ​​4 and 6 for the consecutive numbers 8 and 9, so, being increasing, it cannot take the value 5 for any natural number $n$. (The pedant would say so: $n\leqslant 8\Rightarrow f(n)\leqslant 4;\;\;n\geqslant 9\Rightarrow f(n)\geqslant 6$  and there are no other possibilities.  )

The function $f$ is not surjective.

$\blacksquare$

            Remark CiP

          I suspected from the beginning that the values ​​of the function $f$ make "jumps". I looked for values ​​of $n$ for which each of the three fractions would be a natural number. For this we set the conditions:

$$\begin {cases}n=3\cdot k\;\;\;\;\;\;\;\;(1)\\n+1=5\cdot l\;\;\;(2)\\n+2=7\cdot m\;\;(3)\end{cases}$$

We substitute (1) into (2) and we obtain

$$3\cdot k+1=5\cdot l\;\;\Leftrightarrow\;\;5\cdot l-3\cdot k=1 \tag{kl}$$ 

which is a linear Diophantine equation. We see the solution $k=3,\;l=2$ of (kl) so it has the general solution

$$k=5\cdot p+3,\;\;l=3\cdot p+2 \;,\;\;p\in\mathbb{N}\tag{4}$$

Then  $n\underset{(1)}{=}3\cdot k\underset{(4)}{=}3(5p+3)=15\cdot p+9$ which substituted into (3) gives us

$$(15\cdot p+9)+2=7\cdot m\;\;\Leftrightarrow\;\;7\cdot m-15\cdot p=11 \tag{mp}$$

Here we see an solution  $m=-7,\;p=-4$, so the general solution of  (mp) is

$$m=15\cdot t-7,\;\;p=7\cdot t-4\;,\;\;t\in\mathbb{N} \tag{5}$$

Finally  $n=15p+9\underset{(5)}{=}15(7\cdot t-4)+9=105\cdot t-51.$

     The first value of  $n$ we are looking for is $n=105-51=54.$ Calculating, we get

$f(53)=\left [\frac{53}{3}\right ]+\left [ \frac{54}{5}\right ]+\left [\frac{55}{7}\right ]=\left [17\frac{2}{3}\right ]+\left [10\frac{4}{5}\right ]+\left [ 7\frac{6}{7}\right ]=17+10+7=34,$

$f(54)=\left [\frac{54}{3}\right ]+\left [\frac{55}{5}\right ]+\left [\frac{56}{7}\right ]=18+11+8=37,$

$f(55)=\left [\frac{55}{3}\right ]+\left [\frac{56}{5}\right ]+\left [\frac{57}{7}\right]=\left[18\frac{1}{3}\right]+\left [11\frac{1}{5}\right]+\left[8\frac{1}{7}\right]=18+11+8=37.$

From here it is immediately clear that the function  $f$  is neither injective nor surjective.

<end Rem>

          A PEDANT's remark  Instead of (1)-(3) we could use congruences according to different moduli. 

$n\equiv 0\;(mod\;3),\;\;n+1\equiv 0\;(mod\;5)\Leftrightarrow n\equiv  -1\;(mod\;5),\;\;n\equiv -2\;(mod\;7).$

 Now,  $3\cdot k \equiv -1\;(mod\;5)\Rightarrow 6\cdot k \equiv -2\;(mod\;5)\Leftrightarrow k\equiv -2\;(mod\;5).$ So  $k=5\cdot p-2,\;p\in\mathbb{Z}$  and further  $n=3\cdot k=15\cdot p-6\equiv -2\;(mod\;7)\Rightarrow 15\cdot p\equiv 4\;(mod\;7)\Rightarrow p\equiv 4\equiv -3\;(mod\;7).$ Thus  $p=7\cdot t-3,\;t\in\mathbb{Z}$, so $n=15p-6=15(7t-3)-6=105\cdot t-51.$

<end rem>