Question 46

• remainder $0$:
  The numbers $4$, $8$, $12$, $16$, $20$, .... can be bought by purchasing the correct number of $4$-packs.
• remainder $1$:
  The numbers $9$, $13$, $17$, $21$, $25$, .... can be bought by purchasing one $9$-pack, and the correct number of $4$-packs. We cannot buy $5$ nuggets or $1$ nugget.
• remainder $2$:
  The numbers $18$, $22$, $26$, $30$, $34$, .... can be bought by purchasing two $9$-packs, and the correct number of $4$-packs. We cannot buy $14$, $10$, $6$, or $2$ nuggets.
• remainder $3$:
  The numbers $27$, $31$, $35$, $39$, $43$, .... can be bought by purchasing three $9$-packs, and the correct number of $4$-packs. We cannot buy $23$, $19$, $15$, $11$, $7$, or $3$ nuggets.

In $\bigtriangleup$AOC, X and Z are midpoints of [AO] and [OC] respectively.
$\therefore\quad$[XZ]$\,||\,$[AC] and XZ $=\tfrac{1}{2}$AC $\{$midpoint theorem$\}$

Similarly, in $\bigtriangleup$ABC,
  [MK]$\,||\,$[AC] and MK $=\tfrac{1}{2}$AC

Thus [XZ]$\,||\,$[MK] and XZ $=$ MK
$\therefore\quad$XZKM is a parallelogram.
$\therefore\quad$the diagonals [MZ] and [KX] bisect each other.

It can similarly be shown that LKYX is a parallelogram, and that [LY] bisects [KX].
$\therefore\quad$[KX], [MZ], and [LY] all pass through the midpoint of [KX].
$\therefore\quad$[KX], [MZ], and [LY] are concurrent.