Ce? se întâmplă când are toată familia febră ?? // What happens when the whole family has a fever?

          It seems that the Mathematical Olympiad has begun.

          Let's try to put ourselves in the shoes of a 5th grader. Let's take a list of issues given not too long ago. For example, let's consult the magazine below. 

(Unfortunately I can't show you more than that.)

- The picture is courtesy of SSMR -

          We will choose three 5th grade problems from the first stage of the Olympiad as examples.

          Problem 1 (at page 1) "A test has 20 questions, which are to be solved in 150

                       minutes. The questions are of three types: easy, medium and difficult. 

                       It is expected that each of the 7 easy questions can be solved in 4 

                      minutes and each of the medium questions can be solved in 8 minutes.

                       How many minutes are left for solving difficult questions?"      

                               A. 56            B. 58            C.62            D. 66            E. 70

Answer CiP        D. 66

Solution CiP

               Easy questions and medium questions require a solving time of $7 \cdot 4+7\cdot 8=24$  minutes. So for the difficult questions, $150-84=66$ minutes remain. The answer is D.

$\blacksquare$

               Remark CiP We have 7 easy questions and 7 medium questions, so the number of difficult questions is $20-7-7=6$. So each difficult question is expected to be solved in 11 minutes.

< end Rem>

          In light of the statement in Problem 1, it seems that Problems 1-7 are considered easy, Problems 8-14 are medium, and Problems 15-20 are difficult. I will take one of each. 

           I also mention that the Magazine in the image above does not contain the corresponding answers to these questions. Despite the other qualities: e.g. a clear English that would make even Shakespeare envious. News about the adventures at this Olympics also appeared in the press. That's also where I found out where you can find the Answers. The same lovely SSMR.

          Problem 11 (at page 2) "The largest integer which divided by 2022 gives a

                    quotient smaller than the remainder is :

A. 1048043482         B. 4086461         C. 16185         D. 8091         E. 9         " 

 

Answer CiP    B. 4086461

Solution CiP

               By dividing a number $D$ by $2022$ we obtain the quotient $Q$ and the remainder $R$, according to the long division as below

$$\begin{array}{c|c} D & 2022 \\ \hline \vdots & Q \\ \hline R & \; \end{array}$$

and the Division with remainder Theorem is written

$$D=2022\cdot Q+R\;\; \quad R<2022 \tag{1}$$

The statement also specifies the condition 

$$Q\;<\;R \tag{2}$$

So the first relation (1) gives us $D<2022R+R$ , or $D<2023R$. But from the second relation (1) we  have $D\leqslant 2021$, and so

$$D<2023 \cdot 2021 =4\;088\; 483.$$

So the answer "A."  is excluded, being a larger number.

     We see from (1) that the largest admissible number $D$ occurs when $R$ and $Q$ are the largest, also verifying the condition (2), so 

$$Q=R-1=2020.$$

Hence $D=2022 \cdot 2020+2021=4\;086\;461$, so answer is B. 

$\blacksquare$

           Remark CiP  Answer options C and E also verify the condition (2)

$$16\;185 \;:\;2022 \;=\;8\quad remainder\;9\;\quad \quad 9\;:\;2022\;=\;0\quad remainer\;9$$

but option D does not $8\;091\;;\;2022\;=\;4\quad remainder\;3.$

<end Rem>

          Problem 18 (at page 3) "If $S(n)$ denotes the sum of the digits of the number

                  $n$, then the number of $n-$s so that $\;2021\leqslant n+S(n)\leqslant 2022$ is:

A. 0         B. 1         C. 2         D. 3         E. 4    "

Answer CiP   C. 2

$$1996+S(1996)=1996+25=2021\;\quad \; 2014+S(2014)=2014+7=2021$$

Solution CiP

          $n=0$ does not satisfy the condition.

           If $n\leqslant 1999$ we have $S(n)\leqslant 28$, and then

$$n\geqslant 2021-S(n)\geqslant 2021-28=1993.$$

We have to try the numbers:

$$1993,\quad 1994,\quad\cdots,1999.\quad  \tag{1}$$

After some calculations:

$$1993+S(1993)=1993+22=2015,$$

$$1994+S(1994)=1994+23=2017,$$

$$1995+S(1995)=1995+24=2019,$$

$1996+S(1996)=1996+25=2021$ - this number is convenient

we see that for the other numbers in the list (1) the value $n+S(n)$ exceeds 2023.

          If $n\geqslant 2000$ (anyway $n\leqslant 2022$) we have $S(n)\geqslant 2$ so

 $n\leqslant 2022-S(n)\leqslant 2020$.

Among the numbers   2000, 2001, ..2020 the highest $S(n)$ value is $S(2019)=12$ and then, from $n\geqslant 2021-S(n)$ we get $n\geqslant 2021-12=2009$, so we have to try only the numbers

$$2009,\quad 2010,\;\dots\;2020.\quad \tag{2}$$

Observing the evolution of calculations:

$$2009+S(2009)=2009+11=2020$$

$$2010+S(2010)=2010+3=2013$$

$$2011+S(2011)=2011+4=2015$$

$$2012+S(2012)=2012+5=2017$$

$$2013+S(2013)=2013+6=2019$$

$2014+S(2014)=2014+7=2021$ - this number is convenient

$$2015+S(2015)=2015+8=2023$$

it turns out that from here on out, there can be no more convenient numbers in list (2).

$\blacksquare$