Math Olympiad Problems for Grades 5 and 6: Very Difficult Problems with Solutions

Solution:
The number is [tex]ABC2[/tex].
The sum condition: [tex]A + B + C + 2 = 3C \Rightarrow A + B + 2 = 2C[/tex].
Divisibility by [tex]4[/tex] means ”[tex]C2[/tex]” must be divisible by [tex]4[/tex].
Checking digits for [tex]C[/tex]: [tex]12, 32, 52, 72, 92[/tex] are divisible by [tex]4[/tex].
So [tex]C[/tex] can be [tex]1, 3, 5, 7, 9[/tex].
Divisibility by [tex]9[/tex] means the sum ([tex]3C[/tex]) is a multiple of [tex]9[/tex], so [tex]C[/tex] must be a multiple of [tex]3[/tex].
Thus, [tex]C[/tex] can only be [tex]3[/tex] or [tex]9[/tex].
To maximize [tex]ABCD[/tex], we try [tex]C = 9[/tex].
From [tex]A + B + 2 = 2(9) = 18 \Rightarrow A + B = 16[/tex].
To maximize [tex]A[/tex], let [tex]A = 9[/tex], then [tex]B = 7[/tex].
The number is [tex]9792[/tex].

Solution:
The area of one square face is [tex]4 \times 4 = 16 \text{ cm}^2[/tex].
Two separated cubes have [tex]6 + 6 = 12[/tex] faces.
Joined at one face, [tex]2[/tex] faces are hidden, leaving [tex]12 - 2 = 10[/tex] exposed faces.
The area of the single net is exactly the surface area of the composite figure.
Total Area [tex]= 10 \text{ faces} \times 16 \frac{\text{cm}^2}{\text{face}} = 160 \text{ cm}^2[/tex].

Solution:
Start from the intersection of all three: [tex]3[/tex] students.
Soccer/Basketball only: [tex]10 - 3 = 7[/tex].
Soccer/Volleyball only: [tex]6 - 3 = 3[/tex].
Basketball/Volleyball only: [tex]4 - 3 = 1[/tex].
Soccer only: [tex]24 - (7 + 3 + 3) = 11[/tex].
Basketball only: [tex]19 - (7 + 3 + 1) = 8[/tex].
Volleyball only: [tex]15 - (3 + 3 + 1) = 8[/tex].
Total playing sports: [tex]11 + 7 + 3 + 3 + 8 + 1 + 8 = 41[/tex].
Students playing none: [tex]50 - 41 = 9[/tex].

Solution:
Let [tex]C[/tex] be the total capacity.
Initial volume is [tex]\frac{5}{8}C[/tex].
Half is emptied, meaning [tex]\frac{1}{2} \times \frac{5}{8}C = \frac{5}{16}C[/tex] remains.
Adding [tex]12[/tex] liters makes it [tex]\frac{1}{2}C[/tex].
Equation: [tex]\frac{5}{16}C + 12 = \frac{1}{2}C[/tex].
Subtracting [tex]\frac{5}{16}C[/tex] from both sides gives [tex]12 = \frac{1}{2}C - \frac{5}{16}C = \frac{8}{16}C - \frac{5}{16}C = \frac{3}{16}C[/tex].
Therefore, [tex]C = \frac{12 \times 16}{3} = 4 \times 16 = 64[/tex] liters.

Solution:
The outer perimeter consists exclusively of the two shorter sides of the rectangle and the four equal sides of the two identical isosceles triangles.
The longer sides of the rectangle ([tex]12[/tex] cm) are internal and do not count.
Perimeter [tex]= 2 \times[/tex] (Rectangle shorter side) [tex]+ 4 \times[/tex] (Triangle side) [tex]= 2(8 \text{ cm}) + 4(10 \text{ cm}) = 16 \text{ cm} + 40 \text{ cm} = 56 \text{ cm}[/tex].