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Let $k$ be an integer satisfying $0\le k \le 9\,$. Show that $0\le 10kk^2\le 25$.
Mal from Bedford School in the UK and Mahdi from Mahatma Gandhi International School in India tried all the possible values of $k.$ Here is Mal's work:
Nikita from Jersey College for Girls, Mohamed S from LAE Tottenham in the UK and Sanika from PSBBMS, OMR in India used a graph. The sketch is from Nikita's work. This is how Sanika found the roots and the maximum value:
$y=f(x)=x(10x)$
The roots of the equation are 0 and 10.
Since the squared term has a negative coefficient the parabola will face downwards.
The highest ycoordinate will therefore be in the first quadrant with an x value right between the two x intercepts. (As quadratic equations have symmetric graphs). This implies that the x coordinate is (0+10)$\div$2 = 5.
Substituting this value into the function $f(x)$ will give us the y coordinate; this will be the highest output of the function.
$\therefore 25\geq k(10k)$
If the condition that $0\leq k\leq9$ is considered, then it can be concluded that the lowest ycoordinate is obtained when the xcoordinate is $0$ $\therefore 0\leq k(10k)$
Nikita found the maximum value by completing the square:
$y=(k5)^2+25$
so max point is at $(5,25)$
Mohamed S found the maximum value using differentiation:
$f'(k) = 102k$
To find maximum:
$102k=0$
$10=2k$
$k=5$
$10\times5  5^2 = 25$
$25$ is the maximum
Wiktor from LAE Tottenham and David from the UK proved the inequality algebraically. This is David's work:
Sanika also proved the inequality using the arithmetic mean  geometric mean inequality:
AMGM inequality: $\dfrac{a+b}2 \geq \sqrt{ab}$
Applying AMGM to the numbers $k$ and $(10k)$ we get the following: $$\frac{k+(10k)}2 \geq \sqrt{k(10k)} \Rightarrow 25\geq k(10k)$$The expression will always give a positive integer as output if $0\lt k \leq 9\,.$ So the lowest value can be obtained when $k=0\, .$
Show also that $k(k1)(k+1)$ is divisible by $3\,$.
Mal checked this for the integers 1 to 9:
David, Wiktor, Mohamed S, Mahdi and Nikita wrote this simple proof (this is Wiktor's work):
Nikita and Sanika considered different cases. This is Nikita's work:
For each $3$digit number $N$, where $N\ge100$, let $S$ be the sum of the hundreds digit, the square of the tens digit and the cube of the units digit. Find the numbers $N$ such that $S=N$.
Nikita, David, Mal, Wiktor, Mohamed S, Sanika and Mahdi found an equation linking $a,b$ and $c\,.$ This is Nikita's work:
Nikita then found the possible values of $c^3c$ and went through the values of $a$ to find solutions.
Wiktor and David used part 1 to find out more about $b$. This is Wiktor's work:
Mohamed wrote:
This means that a multiple of $99$ minus a number between $0$ and $25$ must be equal to $c(c+1)(c1).$ $99a$ must be close to $c(c+1)(c1)$ but smaller.
Mahdi, Mohamed, Wiktor and Mal used this idea to go through the possible values of $c$. This is Mahdi's work:
Using all the information, Sanika made a table showing the possible values of $a, b$ and $c$:
All possible values of: 
All possible values of: 
All possible values of: 

$99a$ 
$a$ 
$b(10b)$ 
$b$ 
$c(c1)(c+1)$ 
$c$ 
99 
1 
0 
0 or 10 
120 
5 
198 
2 
9 
1 or 9 
210 
6 
297 
3 
21 
3 or7 
336 
7 
396 
4 
24 
4 or6 
504 
8 
495 
5 


720 
9 
594 
6 




693 
7 




After a couple of combinations, 4 numbers can be derived:
175
135
518
598