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).
$\blacksquare$